Example: f(x) = x 3. 1, x > 0 0, x 0. Example: g(x) =
|
|
- Bennett Norton
- 6 years ago
- Views:
Transcription
1 2.1 Instantaneous rate of cange, or, an informal introduction to derivatives Let a, b be two different values in te domain of f. Te average rate of cange of f between a and b is f(b) f(a) b a. Geometrically, a secant line of te grap of a function f is a line tat passes troug two different points on te grap (a, f(a)) and (b, f(b)). Te slope of te secant line is te average rate of cange of f between a and b. If, as b approaces a, te average rate of cange between a and b approac a number d, ten d is called te instantaneous rate of cange, or te derivative of f at a. Geometrically, wen we fix one of te two points, say fixing a, and let b approaces a, if te secant line converges to a line l a, ten l a is called te tangent line of te grap of f at point (a, f(a)). Te slope of te tangent line is te instantaneous rate of cange. Example: f(x) = x 3 Example: g(x) = { 1, x > 0 0, x Limit Laws lim x c f(x) only depends on te value of f on a open interval containing c excluding c. lim x c C = C, were C is a constant. lim x c x = c Wen lim x c f(x) and lim x c g(x) bot exist, lim x c (f+g)(x) = lim x c f(x)+lim x c g(x), lim x c (f g)(x) = lim x c f(x) lim x c g(x) and lim x c (fg)(x) = lim x c f(x) lim x c g(x). As a consequence, if P is a polynomial, lim x c P (x) = P (c). Wen lim x c f(x) and lim x c g(x) bot exist, and te latter is non-zero, lim x c (f/g)(x) = lim x c f(x) lim x c g(x). As a consequence, if P, Q are bot polynomial and Q(c) 0, lim x c(p/q)(x) = P (c)/q(c). x Example: lim 2 1 x 1 x 3 1 = lim x+1 x 1 x 2 +x+1 = 2/3. Wen lim x c f(x) exists, lim x c f n (x) = (lim x c f(x)) n, lim x c f 1 2n+1 (x) = (limx c f(x)) 1 2n+1. 1
2 2 Wen lim x c f(x) > 0, lim x c f 1 2n (x) = (lim x c f(x)) 1 2n. If f g, lim x c f(x) and lim x c g(x) bot exist, ten lim x c f(x) lim x c g(x). (note: we can t replace wit < ere.) Sandwic teorem: if f g F, lim x c f(x) = lim x c F (x) = L, ten lim x c g(x) = L. Example: lim x 0 sin x x = 1. Classification of te cases wen te limit does not exist: 1: Unbounded 2: Bounded but jumps (i.e. bot left & rigt and limit exist but tey are different.) 3: All remaining cases. We call tem oscillate too muc. 2.3 Te definition of limit Suppose f is defined on an open neigborood of a except possibly at a. By lim x a f(x) = L, we mean ɛ > 0, δ > 0 suc tat x, 0 < x a < δ implies f(x) L < ɛ. ( : for all : tere exists) Example: lim x c x = c [Let δ = ɛ.] As a consequence, by lim x a f(x) does not exist, we mean ɛ > 0, δ > 0, x 1, x 2, 0 < x 1 a < δ, 0 < x 2 a < δ and f(x 1 ) f(x 2 ) > ɛ. In particular, if f jumps at a, lim x a f(x) does not exist. Te proof of some limit laws from definition: Example 1: prove tat lim x c f(x)g(x) = (lim x c f(x))(lim x c g(x)). Let A = lim x c f(x), B = lim x c g(x). ɛ > 0, coose ɛ 1 > 0 suc tat ɛ ɛ 1( A + B ) < ɛ, coose δ > 0 suc tat 0 < x c < δ implies f(x) A < ɛ 1 and g(x) B < ɛ 1, ten 0 < x c < δ implies f(x)g(x) AB < ɛ. Example 2: prove te Sandwic teorem: f g F, lim x c f(x) = lim x c F (x) = L, ten lim x c g(x) exists and equals L.
3 ɛ > 0, coose δ > 0 suc tat 0 < x c < δ implies f(x) L < ɛ and F (x) L < ɛ, ten 0 < x c < δ implies g(x) L < ɛ. Exercises on te concept of limit: 1. lim x 0 sin 1 x does not exist because it (a) is unbounded. (b) jumps. (c) oscillates too muc. [c] 3 2. lim x 0 x sin 1 x (a) does not exist. (b) =0. (c) =1. [a, by Sandwic teorem.] 3. Calculate lim x 0 1+x 1 sin x.[1/2] 4. Calculate lim x 0 (ln( 1 x + 1) ln( 1 x 1)). [0] 5. Calculate lim x 0 sin(1+x) sin(1 x) x. [2 cos 1] 2.4 One-sided limit lim x c f(x) = L if ɛ > 0, δ > 0 suc tat x (c δ, c) = f(x) L < ɛ. lim x c + f(x) = L if ɛ > 0, δ > 0 suc tat x (c, c + δ) = f(x) L < ɛ. Relation wit limit: If lim x c f(x) = L, ten lim x c f(x) = L and lim x c + f(x) = L. If lim x c f(x) = lim x c + f(x) = L, ten lim x c f(x) = L If lim x c f(x) lim x c + f(x) = L, ten lim x c f(x) does not exist. (And f jumps at c.) All te basic laws of limit (e.g. one-sided limit. te Sandwic teorem) can be adapted to work on sin x We can use one-sided limit to sow lim x 0 x = 1 as follows: Step 1: If f is even, lim x 0 + f(x) = lim x 0 f(x) = lim x 0 f(x) Step 2: sin x x is even.
4 4 Step 3: Sow tat lim x 0 + sin x x = 1 wit a geometric argument and te Sandwic teorem. 2.5 Continuity We call f continuous at c if lim x c f(x) = f(c), left continuous if lim x c f(x) = f(c), rigt continuous if lim x c + f(x) = f(c). f is called continuous if it is continuous everywere in its domain. Geometric meaning: if f is continuous on interval I, ten te grap of f restricted on I is connected. Properties: If f is continuous at c, lim x a g(x) = c, ten lim x a f(g(x)) = f(c). Specifically, te composition of continuous functions are continuous. Rational functions, x a, sin x, cos x, e x, ln x are all continuous. Sum, difference, product, quotient, power, roots preserve continuity. Intermediate Value Teorem: if f is continuous on interval I, a, b I suc tat f(a) < c < f(b), ten tere exists x I suc tat f(x) = c. Examples: (1) Find all x suc tat f(x) = [x] (floor of x) is continuous, left continuous, rigt continuous. (2) Sow tat cot x is continuous. (3) True or false: you were once exactly 3 ft tall. (use Intermediate Value Teorem) (4) We can approximate e 2 wit because (a) e x is continuous (b) x 2 is continuous. ((b), because x 2 is continuous implies tat x 2 is close to x 2 if x is close to x.) { x, x Q (5) Find all x were f(x) = is continuous. (0) 0, oterwise Furter exercises: (1) Sow tat f(x) = x is continuous. (Hint: use definition to sow continuity at 0.) (2) Sow tat polynomial x x x 2 + x as a real root. (Hint: use Intermediate Value Teorem.)
5 Continuous extension: If f is not defined on c, but lim { x c f(x) exists, we can extend f(x), x D(f) te function f to c by defining a new function F (x) =. Wen lim x c f(x), x = c defining exponential functions we used te idea of continuous extension. (3) Assuming te temperature distribution is continuous, sow tat tere are 2 diametrically opposite points on te equator wit te same temperature at te same time. (Hint: Intermediate Value Teorem.) 5 Answers to te problems in te Quantifier workseet (b) f is bounded on any finite interval. Example of a function { tat satisfies (b): 0, x = 0 f 1 (x) = x. Example of a function tat doesn t satisfy (b): g 1 (x) = 1 x, x 0. (c) f is constant on an open interval containing x 0. Example of a function tat satisfies (b): f 2 (x) = 0. Example of a function tat doesn t satisfy (b): g 2 (x) = x. (d) f is bounded. Example of a function tat satisfies (b): f 3 (x) = sin x. Example of a function tat doesn t satisfy (b): g 3 (x) = x. (e) Te preimage of any finite interval under f is bounded. Example of a function tat satisfies (e): f 4 (x) = x. Example of a function tat doesn t satisfy (b): g 4 (x) = sin x. 2.6 Limits Involving Infinity We say lim x c f(x) = L, if f(x) can be arbitrary close to L for all x c tat are sufficiently close to c. (Tis is anoter way of stating te ɛ δ definition.) We say lim x f(x) = L, if f(x) can be arbitrary close to L for all x tat are sufficiently large. We say lim x f(x) = L, if f(x) can be arbitrary close to L for all x suc tat x are sufficiently large. All limit laws we learned earlier are true for lim x and lim x. Example: lim x (ln(x + 1) ln(x 1)). We say lim x c f(x) =, if f(x) can be arbitrary large for all x c tat are sufficiently close to c.
6 6 We can define infinite one-sided limit, infinite limit as x ± similarly. Example: Calculate lim x 0 + tan 1 ( 1 x ). Here tan 1 is te arctangent function, wic is te inverse of f(x) = tan(x), π/2 < x < π/2. Horizontal, Oblique & Vertical Asymptotes: orizontal & vertical asymptotes are just ways of saying te existence of finite limit as x goes to infinity, or te existence of infinite limit as x approac some finite number. We say te grap of f as an oblique asymptote y = kx + b, if f(x) kx b as limit 0 as x ±. Example: f(x) = x2 +1 x = x + 1 x. Review for Capter 1 & 2 Definition of function, domain, range, natural domain Definition of odd/even/increasing/decreasing functions Grap of a function Operations on functions: sum, difference, product, quotients, composition Te sifting and scaling of te grap of a function Definition of trigonometric functions, exponential functions and logaritms, identities involving tese functions One-to-one functions and teir inverse Definition of inverse trigonometric functions Limit laws, Sandwic Teorem sin x lim x 0 x = 1 Classification of te cases were limit does not exist: unbounded, jumps and oscillates too muc Relationsip between limit and one-sided limits Continuity and Intermediate Value Teorem Limit involving infinity Horizontal, Oblique & Vertical Asymptotes Average and Instantaneous rate of cange Review Questions: 1. Wat is te relationsip between te grap of tan(2x) and tan(x)? Write tan(2x) in terms of sin(x) and cos(x), and calculate lim x π/3 tan(2x). 2. f(x) = sin 1 x is an (a) even function. (b) odd function. (c) increasing function. (d) decreasing function. 3. Sow tat tere is a x suc tat x 2 = e x.
7 7 { b, x = 0 4. Find b suc tat f(x) = e 1 is continuous. x 2, x 0
8 8 Trigonometric functions & identities: Te unit circle is te circle centered at origin wit radius 1. (1, 0) In matematics, we measure angles by radian, i.e. te lengt of te arc on te unit circle tat subtend te angle from te origin. Given any real number s, consider te arc on te unit circle starting at (1, 0) and as lengt s. Te y-coordinate and x-coordinate of tis arc is defined as sin s and cos s. By definition, sin is odd and cos is even, bot ave period 2π, and sin 2 x + cos 2 x = 1. Oter trigonometric functions are defined as follows: tan x = sin x cos x sec x = 1 cos x, csc x = 1 sin x., cot x = cos x sin x, Oter trigonometric identities you need to remember: sin(a+b) = sin a cos b+sin b cos a, cos(a + b) = cos a cos b sin a sin b, cos 2x = cos 2 x sin 2 x = 2 cos 2 x 1 = 1 2 sin 2 x, sin 2x = 2 sin x cos x. Inverse functions: If for any a, b D(f), a b implies f(a) f(b), we call f oneto-one. If f is one-to-one, define function f 1 on f(d(f)) as: f 1 (y) = x if and only if f(x) = y. We ave: for any x D(f), f 1 (f(x)) = x; for any y f(d(f)), f(f 1 (y)) = y. We call f 1 te inverse function of f. To grap f 1, do a reflection of te grap of f by te line x = y. Secant line and tangent line: A secant line of te grap of a function f is a line tat passes troug two points (a, f(a)) and (b, f(b)), suc tat a b. Te slope of te secant line is called te average rate of cange of f between a and b. Wen we fix one of te two points, say fixing a, and let b approaces a, if te secant line converges to a line l a, ten l a is called te tangent line of te grap of f at point (a, f(a)). Te slope of te tangent line is called te instantaneous rate of cange of f at a. Exponential and Logaritm: Exponential functions are functions of te form f(x) = a x, were a is a positive real number. Te number e is defined by te property tat te
9 tangent line of te grap of y = e x at (0, 1) as slope 1. ln x is te inverse function of e x. Relation between limit and continuity f is continuous at c if (i) te limit of f at c exists, and (ii) tis limit equals f(c). Te existence of limit at a point c does not depend on te value of f(c), but continuity at c does. Finding Asymptotes To find orizontal or vertical asymptotes, calculate te limit. To find oblique asymptote of a rational function, write it into te form of ax + b + f(x) suc tat lim x ± f(x) = 0. Graping wit computer, ɛ and δ: Tese are not required topics & will not appear in Prelim 1. Exercises: 1. Write cos(4x) in terms of cos x. How about cos(5x)? 2. Find te inverse function of f(x) = ln(x+1) ln(x 1). Wat is its domain and range? 3. Find te tangent line of y = x at (1, 1). 9
10 10 3.1,3.2 Derivative f (x) = lim 0 f(x + ) f(x) Left & rigt and derivative: lim 0 f(x+) f(x) & lim 0 + f(x+) f(x) Te existence of derivative implies continuity. Te existence of left (rigt) derivative implies left (rigt) continuity. Cases wen derivative not exist: discontinuity, corner, cusp, vertical tangent line, etc. Example: f(x) = 3 x, calculate f. (int: use a 3 b 3 = (a b)(a 2 + ab + b 2 )) Wen te derivate of f exist for all x A, we call f differentiable on A. Example: 1 x Additional Example: f(x) = { 0, x = 0 x 3 sin 1 x, at x = 0. (int: use sandwic teorem) x Differentiation Rules, n-t Derivative Te function f itself is te 0-t derivative of f. Te n-t derivative of f, denoted as f (n), is te derivative of te n 1-t derivative of f. f (1) is often written as f, f (2) as f, f (3) as f etc. Geometric meaning: f increasing implies f > 0. f decreasing implies f < 0. f reaces local maximum or minimum at x implies f (x) = 0. f is concave up (or convex in some books) implies f > 0. f is concave down (or concave in some books) implies f < 0. Example: (sin x) = lim 0 sin cos x+sin x cos sin x cos 1 ( 2 )2 4 ) = cos x + lim 0 (sin x 2 sin2 2 = lim 0 (cos x sin ) + lim 0(sin x ) = cos x + sin x 2 0 = cos x.
11 11 Rules for derivatives a and b are constants, ten (af + bg) = af + bg. (fg) = f g + fg. ( f g ) = f g g f g 2. Cain Rule: (f(g(x))) = g (x)f (g(x)). Inverse function rule: (f 1 (x)) = 1 f (f 1 (x)). (x a ) = ax a 1, (sin x) = cos x, (cos x) = sin x, (e x ) = e x, (ln x) = 1 x, (sin 1 x) = 1 1 x 2, (cos 1 x) = 1, 1 x 2 (tan 1 x) = 1. 1+x 2 Exercises: (tan x) ( ex x 2 +1 ) { 1, x = 0 f(x) = sin x find f x, x 0, Proof tat f (0) = 0: f sin (0) = lim 1 sin sin 0 = lim 0. is an odd function, 2 2 sin so we only need to sow lim 0 + = 0. Use Sandwic Teorem. Wen > 0, 2 sin tan sin sin tan sin tan, so 0. lim = 0, lim 0 + = 2 lim 0 +( sin cos 1 cos ) = lim 0 sin + lim cos lim 2 sin lim ( 0 2 )2 + 4 = 0, so
12 12 sin lim g(x) = (x sin x ) = 0. q.e.d. { 0, x = 0 x sin 1 find g x, x 0,
13 3.4 Application of Derivatives Speed, acceleration, jerk (i.e. te first, second, and tird order derivative of te displacement wit regards to time) Margin in economics. Sensitivity to cange , Proof of te rules of derivatives (i) a and b are constants, ten (af + bg) = af + bg. (ii) (fg) = f g + fg. (iii) ( f g ) = f g g f g 2. (iv) Cain Rule: (f(g(x))) = g (x)f (g(x)). (v) Inverse function rule: (f 1 (x)) = 1 f (f 1 (x)). (vi) (x a ) = ax a 1 (vii) (sin x) = cos x (viii) (cos x) = sin x (ix) (e x ) = e x (x) (ln x) = 1 x (xi) (sin 1 x) = 1 1 x 2 (xii) (cos 1 x) = 1 1 x 2 (xiii) (tan 1 x) = 1 1+x 2. Proof. (i) follows directly from te laws of limit. (ii) (fg) f(x+)g(x+) f(x)g(x) (x) = lim 0 f g + fg. = lim 0 (f(x+)g(x+) f(x)g(x+))+(f(x)g(x+) f(x)g(x)) = (iii) f g = f g 1, ence it follows from (ii), (iv) and (vi). (iv) is based on an equivalent definition of derivative as linear approximation. So, f(g(x + )) = f(g(x) + g (x) + o()) = f(g(x)) + f (g(x))g (x) + f (g(x))o() + o(g (x) + o()) = f(g(x)) + f (g(x))g (x) + o(). (v) follows from (iv) and te fact tat f(f 1 (x)) = x. (vi) for positive x, x a = e ln xa = e a ln x, ence te formula follows from (iv), (ix) and (x). For negative x, use ( x) a = ( 1) a x a.
14 14 (vii) sin sin(x+) sin(x) (x) = lim 0 cos x = cos x. sin x(cos 1) cos x sin = lim 0 +lim 0 = lim 0 sin ( 2 sin 2 2 ) + (viii) follows from (vii) and te fact tat cos(x) = sin(π/2 x). (ix) (e x ) = lim 0 e x+ e x = e x lim 0 e 1 = e x (e x ) x=0 = e x. (x) follows from (ix) and (v). (xi) follows from (v) and (vii). (xii) follows from (v) and (viii). (xiii) follows from (iii), (v), (vii), and (viii). Example: (cot 1 x) ( x a ) (log 1+x 2(1 + x 4 )) 3.11 Te Proof of te Cain Rule Teorem 1. f (x) = a if and only if tere exists a function ɛ defined around 0 suc tat f(x + ) = f(x) + a + ɛ(), and lim 0 ɛ() = 0. Remark 1. Te function ɛ depends on f and x. Proof. Suppose f (x) = a, ten f(x + ) = f(x) + a + f(x + ) f(x) a = f(x) + a + [ f(x+) f(x) f(x+) f(x) a], and lim 0 a = f (x) a = 0, so we can let ɛ() = a wen 0 and ɛ() = 0 wen = 0, and ave f(x+) = f(x)+a+ɛ(). f(x+) f(x) Suppose tere is a function ɛ suc tat f(x+) = f(x)+a+ɛ(), and lim 0 ɛ() = 0, f f(x+) f(x) a+ɛ() (x) = lim 0 = lim 0 = a + lim 0 ɛ() = a. Now we can prove te cain rule using te previous teorem: f(g(x+)) = f(g(x)+g (x)+ɛ g ()) = f(g(x))+f (g(x))(g (x)+ɛ g ())+ɛ f (g (x)+ ɛ g ())(g (x)+ɛ g ()) = f(g(x))+f (g(x))g (x)+[f (g(x))ɛ g ()+ɛ f (g (x)+ɛ g ())(g (x)+ ɛ g ())], were ɛ f is te ɛ function corresponding to f and g(x) and ɛ g is te ɛ function corresponding to g and x. lim 0 [f (g(x))ɛ g () + ɛ f (g (x) + ɛ g ())(g (x) + ɛ g ())] = 0, so (f(g(x))) = f (g(x))g (x).
15 Remark 2. Te geometric meaning of ɛ is te difference between te slope of te secant line and te tangent line. 3.7 Derivative of implicit function Suppose function y(x) satisfies (y(x)) 2 + xy(x) + x 2 = 3, y(1) = 1, find y (1), y (1). (Hint: use Cain Rule) 15
16 16 Proof of te Rules of Differentiation: (1) a and b are constants, ten (af + bg) = af + bg (2) (sin x) = cos x (3) (e x ) = e x (4) (fg) = f g + fg (5) Cain Rule: (f(g(x))) = g (x)f (g(x)) (6) Inverse function rule: (f 1 (x)) 1 = f (f 1 (x)) (7) (cos x) = sin x (8) (ln x) = 1 x (9) (x a ) = ax a 1 (10) ( f g ) = f g g f g 2 (11) (sin 1 x) = 1 1 x 2 (12) (cos 1 x) = 1 1 x 2 (13) (tan 1 x) = 1 1+x 2 Proof. (1) follows directly from te limit laws. (2) sin sin(x+) sin(x) sin x(cos 1) cos x sin sin x( 2 sin (x) = lim 0 = lim 0 +lim 0 = lim 2 2 ) 0 + cos x = 2 sin x lim 0 ( sin 2 ) cos x = cos x. Here, we used te following 2 trigonometric identities: sin(a + b) = sin a cos b + cos a sin b, and cos c = 1 2 sin 2 c 2 2. (3) Let f(x) = e x, ten by te definition of number e, f (0) = 1. f (x) = lim 0 e x+ e x = e x lim 0 e 1 = e x f (0) = e x. (4) (fg) f(x+)g(x+) f(x)g(x) (x) = lim 0 f g + fg. = lim 0 (f(x+)g(x+) f(x)g(x+))+(f(x)g(x+) f(x)g(x)) = (5) Lemma. f (x) = a if and only if tere exists a function ɛ defined around 0 suc tat f(x + ) = f(x) + a + ɛ(), and lim 0 ɛ() = 0. Proof. Suppose f (x) = a, ten f(x + ) = f(x) + a + f(x + ) f(x) a = f(x) + a + [ f(x+) f(x) f(x+) f(x) a], and lim 0 a = f (x) a = 0, so if we define ɛ() = { f(x+) f(x) a, 0, ɛ satisfies all te requirements in te lemma. 0, = 0 On te oter and, suppose tere is a function ɛ suc tat f(x + ) = f(x) + a + ɛ(), and lim 0 ɛ() = 0, ten f f(x+) f(x) a+ɛ() (x) = lim 0 = lim 0 = a + lim 0 ɛ() = a.
17 Remark. Geometrically, te function ɛ is te difference between te slope of te secant line between (x, f(x)) and (x +, f(x + )) and te slope of te tangent line at (x, f(x)), and te lemma is saying tat te existence of limit is equivalent to te convergence of te secant line as 0. ɛ depends on f and x. Now we can prove te cain rule using te previous lemma: f(g(x+)) = f(g(x)+g (x)+ɛ g ()) = f(g(x))+f (g(x))(g (x)+ɛ g ())+ɛ f (g (x)+ ɛ g ())(g (x)+ɛ g ()) = f(g(x))+f (g(x))g (x)+[f (g(x))ɛ g ()+ɛ f (g (x)+ɛ g ())(g (x)+ ɛ g ())], were ɛ f is te ɛ function corresponding to f and g(x) and ɛ g is te ɛ function corresponding to g and x. lim 0 [f (g(x))ɛ g () + ɛ f (g (x) + ɛ g ())(g (x) + ɛ g ())] = 0, so (f(g(x))) = f (g(x))g (x). (6)-(13) follows from (1)-(5). Examples: 1. Function y(x) satisfy x 2 + yx + y 2 = 3, y(1) = 1. Calculate y (1) and y (1). Solution: By assumption, F (x) = x 2 + y(x)x + (y(x)) 2 = 3, so F = F = 0. F = 2x + y + y x + 2y y = 0 so y (1) = 1. F = 2 + y + y x + y + 2y y + 2y y = 0, so 3y + 2 = 0, y = 2/3. (Sorry I made a mistake on tis problem in te morning.) 2. (csc 1 ) Solution 1: If y = csc 1 (x) ten x = csc y = 1 sin y, so sin y = 1 x, y = sin 1 1 x. Now use te cain rule, (csc 1 (x)) = (sin 1 1 x ) = 1 x 2 1 = 1 1 x, and x 2 1 (csc 1 (x)) = ( x x 2 1) = x 2 (x 2 1) x 2 x x x 2 1+ x x x 2 1 = x (2x2 1) x 2 (x 2 1) x 3 (x 2 1) 2 3. Here we do not need to deal wit x = 0 because 0 is not in te domain of csc 1. Solution 2: Use te inverse rule, (csc 1 (x)) = 1 csc (csc 1 x) = 1 cos(csc 1 x) sin 2 (csc 1 x) 17 = sin2 (csc 1 x). 1 sin(csc 1 x) Because if u = csc 1 (x) ten x = 1 sin u, so sin(csc 1 x) = sin u = 1 x, (csc 1 (x)) = 1 x 2 Ten do te same calculation as in Solution (tan 1 (ln(x))) Solution: By cain rule, tan 1 (ln(x))) = 1 x 1+(ln(x)) x 2
18 18 Oter topics on derivative (1) Te normal line of a plane curve γ troug a point P on γ is te line passes troug P and is ortogonal to te tangent line of γ at P. Example: Find te tangent line and normal line of y 2 + xy + x 2 = 3 at (1, 1). Example: Exercise 52 (2) Calculate te composition of trigonometric and inverse trigonometric function: Example: cos(sec 1 (x)): if y = sec 1 x, ten x = sec y = 1 cos y, so sec is te composition of 1 x and cos. We know tat (f g) 1 = g 1 f 1, so sec 1 (x) = cos( 1 x ), cos(sec 1 (x)) = 1 x. Example: tan(sin 1 x). (3) Calculate te derivative of products. ( i f i) = ( f i i f i ) i f i. (proved by induction or by applying ln on bot sides) Calculate ((1 + x 2 )(1 + x 4 )(1 + x 6 )(1 + x 8 )) (4) Use derivative to calculate limit: Calculate lim x 0 (1 + x) 1 x Calculate lim x 0 sin x sin 3x Homework from , : te slope of te tangent line is te derivative : te rate of cange is te derivative. { x 2 sin(1/x), x 0 (*)3.1.35: calculate te derivative of f(x) = at x = 0 by definition. 0, x = , , 3.3.2, 3.3.6, , 3.5.2, 3.5.6, , , 3.6.6, , , , , : calculation of derivative wit te rules : use differentiation rules : maximum iff derivative is 0.
19 19 (*)3.2.28, , : sketcing derivative. (*)3.5.57: use te limit of sin(x)/x. (*)3.7.16, , : implicit differentiation. Teorem 2. e = lim x 0 (1 + x) 1/x Te number e Proof. lim x 0 (1 + x) 1/x = lim x 0 (e ln(1+x) x ) = e lim x 0 ln(1+x) ln(1) x = e (ln) (1) = e. Remark 3. We will see later or in Calc 2 tat e = n 1 n!.
20 20 Word problems wit differentiation Steps: 1. Name te variables. 2. Write down te given numbers (if tere are any). 3. Write down te relation between variables as equations. 4. Differentiate (often w.r.t. time). 5. Find te answer. Example 0: A point moves on te unit circle x 2 + y 2 = 1. If x(t 0 ) = 0.6, y(t 0 ) = 0.8, x (t 0 ) = 1, x (t 0 ) = 0, find y (t 0 ) and y (t 0 ). Example 1: Point A is moving on te positive y-axis downwards wit speed 1, and is currently at (0, 1), point B is moving on te positive x-axis and te distance between tem is always 2. Find te speed of point B and te midpoint of line segment AB. Example 2: (from Kepler s laws to universal gravitation) Te orbit of a planet around its sun is ρ(a + sin θ) = 1, were a > 1. (1) Find te relationsip between dρ dθ dt and dt. (2) If we furter assume ρ 2 dθ d2 dt = 1, write (ρ cos θ) as a function of ρ and θ. dt 2 (3) Do te same for ρ sin θ. (4) Write te acceleration of te planet as a function of ρ and θ. Solution for Example 2: (1) ρ (a + sin θ) + ρθ cos θ = 0 (2) Because ρ 2 θ = 1, ρ (a + sin θ) + cos θ ρ = 0, ρ = cos θ. Hence, (ρ cos θ) = ρ cos θ ρθ sin θ = cos 2 θ sin θ ρ = cos 2 θ (a + sin θ) sin θ = 1 a sin θ, (ρ cos θ) = (1 a sin θ) = a cos θ. ρ 2 (3) Similar to (2), we ave (ρ sin θ) = a ρ 2 sin θ. (4) From (2), (3), we know te acceleration is a ρ 2 in te direction towards te origin Standard linear approximation and differential If f is differentiable at a, we call f(a + ) f(a) + f (a) te standard linear approximation. We can also let x = a + and write it as f(x) f(a) + (x a)f (a).
21 Formula like sin (x) = cos x can be written as d sin x = cos xdx. Here te symbol d followed by a function is called a differential Critical points a is called a critical point of f if f (a) = 0, or a is in te interior of D(f) and f (a) is undefined. Remark 4. Sorry, I made a mistake in class in te definition of critical points. Exercises: 1. Find te domain, range, and derivative of y = ln(ln x). 2. Find te derivative of sin(cot 1 x). { x 2 sin(ln x ), x 0 3. Let f(x) =. Find all its critical points. 0, x = 0 We call a a local maximum of f, if tere is an open interval I containing a suc tat f(a) f(x) for all x in I D(f). Local minimum is defined similarly. Teorem 3. If f is differentiable at a, a is a local maximum or local minimum, ten f (a) = 0. Teorem 4. If f is continuous and is defined on a finite closed interval [a, b], ten it reaces its maximum and minimum somewere on [a, b]. Te above teorem follows from te following property of real numbers: Lemma 5. Let {x n } be an infinite sequence of real numbers in a finite closed interval I, tere must be some y I suc tat any open interval containing y also contains infinitely many x n. Examples: Find te maximum and minimum of te following functions: 1. x x 2 2. e x + e x 4.2, 4.3 Mean Value Teorem and its applications Teorem If f is continuous on [a, b] and differentiable on (a, b), f(a) = f(b), ten tere is some c (a, b) suc tat f (c) = 0.
22 22 2. If f is continuous on [a, b] and differentiable on (a, b), ten tere is some c (a, b) suc tat f (c) = f(b) f(a) b a. Corollary If f (x) = 0 on an interval I ten f is constant on tat interval. 2. If f = g on an interval I ten f g is constant on tat interval. 3. If f > 0 on an interval I ten f is increasing on I, If f < 0 on an interval I ten f is decreasing on I. Examples: 1. Sow tat e x 100x 2 as at most 3 zeros. 2. Sow tat ln(ab) = ln a + ln b { x 2 sin(1/x) + x/10, x 0 3. Consider function f(x) = 0, x = 0. { f 2x sin(1/x) cos(1/x) + 0.1, x 0 (x) = 0.1, x = 0. f (0) > 0 but f is neiter increasing nor decreasing on any interval containing 0. Consider te derivative: 2x sin(1/x) cos(1/x) + 0.1, te first term is small for small x, te second term oscillates between ±1, so te derivative oscillates between 1.1 and 0.9 as x 0. In oter words, any interval I containing x must ave some region were f > 0 and some region were f < 0, ence f can not be increasing or decreasing on I. 4. Sow tat cos x 1 x 2 /2. Because cos and 1 x 2 /2 are bot even we only need to sow it for x > 0. Consider F (x) = cos x (1 x 2 /2). F (x) = sin x + x > 0, so F is increasing wen x > 0. We also know tat F (0) = 0, so F (x) 0 for all x 0, i.e. cos x 1 x 2 /2. Corollary 2. (First derivative test) c is a local min (max) if f canges from positive to negative (negative to positive) at c. Example: f(x) = x 3 x. f = 3x 2 1, f > 0 wen x < 3 3 or x > 3 3 < x < 3 3, so 3 3 is a local max and 3 3 is a local min. 3 3, f < 0 wen Exercises: Find te local max/min of te following functions: 1. x x e x Solution: (x x e x ) = (e x ln x e x ) = (e x ln x x ) = (x ln x x) e x ln x x = (ln x + 1 1)e x ln x x = ln xe x ln x x. So by te first derivative test, x = 1 is a local minimum. 2. x 4 x 2
23 Solution: x 4 x 2 = { x 4 x 2, x 1 or x 1 x 2 x 4, 1 < x < By calculating derivative we know 2 2, 0), ( 2 2, 1), ), and (1, ). So 0, ±1 are te tat f(x) = x 4 x 2 is decreasing on te following intervals: (, 1), ( and increasing on te following intervals: ( 1, local minimums and ± 3. sin(x) + sin(x + 1) are te local maximums. 2 ), (0, 2 2 Solution 1: sin(x) + sin(x + 1) = sin((x + 1/2) 1/2) + sin((x + 1/2) + 1/2) = 2 sin(x + 1/2) cos(1/2). So it reaces local maximum wen x = π kπ, local minimum wen x = π kπ, were k Z. Solution 2: Let f(x) = sin(x) + sin(x + 1), ten f (x) = cos x + cos(x + 1) = cos x + cos x cos 1 sin x sin 1 = (1+cos 1) cos x sin 1 sin x, ence f (x) = 0 means tan x = 1+cos 1 sin 1, i.e. x {x k } were x k = tan 1 1+cos 1 sin 1 + kπ, were k Z. f (x) = (1 + cos 1) cos x sin 1 sin x = cos x sin 1 ( 1+cos 1 sin 1 tan x), so f is positive between x 2k 1 and x 2k, and negative between x 2k and x 2k+1, ence x k is local max wen k is even, local min wen k is odd x5 2 3 x3 + x 1 17 Solution: Let f(x) = 1 5 x5 2 3 x3 + x f (x) = (x 2 1) 2 0, and f (x) > 0 wen x ±1. So, f is increasing on (, 1), ( 1, 1) and (1, ), wic implies tat f is increasing on R, ence it as no local min or local max. concavity and second derivative test f > 0 implies tat f is increasing, wic we call f is concave up. f < 0 implies tat f is decreasing, wic we call f is concave down. Places were tangent line exists (i.e. f exists or f = ) and concavity canges are called inflection points. By Mean Value Teorem (MVT), concave up implies tat any line segment linking two points on te grap of f lies above te grap of f, and concave down implies tat any line segment linking two points on te grap of f lies below te grap of f. In oter mat textbooks concave up is often called convex and concave down is often called concave.
24 24 In te future, wen sketcing te grap of a function, we need to pay attention to concavity. Second derivative test: if f (a) = 0 and f (a) > 0 ten a is local min, if f (a) = 0 and f (a) < 0 ten a is a local max. Example: sketc te grap of x 3 x. Sketc te grap of tan 1 x. Sketc te grap of x x. L Hospital s rule Tis is a generalization of te argument we used earlier to calculate limits like lim x 0 sin x ln(x+1). Teorem If lim x a + f = lim x a + g = 0, g and g are nonzero on some interval (a, a + d), and lim f (x) x a + g (x) = A exist, ten lim x a f(x) + g(x) also exists and equals A. 2. If lim x a + f = lim x a + g =, g and g are nonzero on some interval (a, a + d), and lim f (x) x a + g (x) = A exist, ten lim x a f(x) + g(x) also exists and equals A. Example: lim x 0 cos x 1 x 2. lim x 0 + x x (Hint: rewrite it as e ln x x 1/2. Non-example: Wen f(x) = x 2 sin 1 x, g(x) = x, lim x 0 f(x) g(x) = 0, but lim x 0 f (x) g (x) does not exist. lim x 0 sin x x x 3. (answer: 1 6 ) lim x 0 ( 1 ln(x+1) 1 x ) 1 Solution: lim x 0 ( ln(x+1) 1 x ) = lim x 0 x ln(x+1) 1 x ln(x+1) = lim 1 x+1 x 0 1 lim x 0 1+ln(x+1)+1 = 1 2. ln(x+1)+ x x+1 = lim x 0 x (x+1) ln(x+1)+x = 1 Remark 5. Tere are many ways to write convenient for te application of l Hospital rule. ln(x+1) 1 x into a quotient, but not all are Proof of te L Hospital s rule:
25 Lemma 8. (Caucy s MVT) If f and g are bot continuous on [a, b] and differentiable on (a, b), g(a) g(b), ten tere is a c (a, b) suc tat f (c) g (c) = f(b) f(a) g(b) g(a). Proof of Lemma: Use te MVT on F (x) = f(x) f(a) f(b) f(a) g(b) g(a) (g(x) g(a)). Remark 6. A common tecnique for applying MVT is to build functions ten use MVT on tem. For example, if we know f(0) = f(1) = 1, f 1, ten f(x) 1 8. To sow tat, consider F (x) = f(x) 1 2x(1 x). Now we only need to sow tat F 0. If not, by MVT, tere is some a (0, 1) suc tat F (a) > 0, ence tere is some b (0, a) suc tat F (b) > 0, and some c (a, 1) suc tat F (c) < 0. Use MVT again, tere is some point m (b, c) (0, 1) suc tat F (m) = f (m) ( 1) < 0, a contradiction. It is easy to see tat ere 1 8 is te best possible bound. Now we can prove L Hospital s rule. Te proof of part 1 is in te textbook. Te proof of part 2 is as follows: Use te definition of limit. Given ɛ > 0. Because te limit of f /g is A, tere is some δ > 0 suc tat on (a, a + δ), f /g A < ɛ/2. Let M = 1000 ɛ max{ f(a + δ), g(a + δ) }, 0 < δ < δ suc tat on (a, a+δ ), g > 2 ɛ max{ g(a + δ), f(a + δ) (A ɛ)g(a + δ), f(a + δ) (A + ɛ)g(a + δ) }. Now for any x (a, a + δ ), use Caucy s MVT on (a, a + δ), we ave A ɛ 2 < f(x) f(a+δ) g(x) g(a+δ) < A + ɛ 2. Hence, f(x) f(a + δ) = A (g(x) g(a + δ)), were A A < ɛ/2. Terefore, f(x) g(x) = A + { f(a+δ) A g(a+δ) g(x) }, and te assumption on δ tells us tat te term in {} is bounded by ɛ 2. q.e.d. Summary on sketcing te grap of a function Wen sketcing te grap of a function, pay attention to te following: domain, symmetry, critical points, limit points, boundary points, monotonicity, inflection points, concavity, asymptotes, intersection wit x and y axis. Wen te exact positions of certain points can not be written down explicitly, sketc its approximated position. Example: sketc te grap of e 1 x Applications of max & min Example 1: Find te maximum volume of rigt circular cones wit surface area S. Solution: A rigt circular cone wit radius r and eigt as volume V = 1 3 πr2 and surface area S = πr r πr = πr(r + r ). Fix S, solve for wit respect to r we get = 1 r S π S π 2r2, ence V = 1 3 πr2 = 1 3 πs r S π 2r2. Use te first
26 26 derivative test we know tat V reaces its maximum wen r = 1 2 S is π 1 2 Example 2: (maximum sustainable yield) ky(c y) M = 0, maximize M. S π, and te maximum Example 3: (te dual of Example 1) Find te minimal surface area of rigt circular cones wit volume V. Solution: A rigt circular cone wit radius r and eigt as volume V = 1 3 πr2 and surface area S = πr r πr = πr(r + r ). Fix V, solve for wit respect to r we get = 3V, ence S = πr(r + r πr ) = πr(r + r 2 + 9V 2 ). Use te first 2 π 2 r 4 derivative test we know tat S reaces its minimum wen r = ( 9V 2 ) 1 8π 2 6, and te minimum is π 1 3 V 2 3. x n+1 = x n f(x n )/f (x n ) 4.7 Newton s Metod Situations in wic it converges slowly or does not converge: non-existence or noncontinuous derivative, te starting point is too far away, f (x ) = 0 were x is te root we are trying to find, etc. 4.8 Antiderivative F = f, ten F is called an antiderivative of f. Te collection of all antiderivatives of f is called te indefinite integral of f: f(x)dx. By MVT, if f is defined on an interval, two antiderivatives of f must differ by a constant. So, if F = f ten fdx = F + C. Example: (x + 1)dx, sec x tan x + e x dx, sin 2x + cos xdx. Terminal velocity: if dv dt = g kv2, ten t = 1 g kv 2 dv. Unlike differentiation, it is ard to find te antiderivative of elementary function symbolically. Te computer algebra system Maxima implemented an algoritm for symbolic integration. You can read te source code (in Common Lisp) ere: ttp://sourceforge.net/p/maxima/code/ci/master/tree/src/risc.lisp
27 Area, upper/lower/midpoint sums Te concept of integration is based on our intuition about area. Te problem of finding displacement from velocity, and finding te average of a function, can be reduced to finding area. To estimate te area below a function f, wic is defined on a finite interval [a, b], we can decompose [a, b] into n subintervals of equal lengt, and define te following concepts: b a (1) Upper sum: n 1 n k=0 M k, were M k is te maximum of f on [a + k(b a) n, a + ]. Wen te maximum does not exist, use te smallest upper bound. (k+1)(b a) n (2) Lower sum: b a (k+1)(b a) n (3) Mid-point sum: b a n n 1 n k=0 m k, were m k is te minimum of f on [a + k(b a) ]. Wen te minimum does not exist, use te greatest lower bound. (4) Left-end-point sum: b a n n 1 n 1 k=0 (5) Rigt-end-point sum: b a n k=0 f(a + (k+ 1 2 )(b a) n ). k(b a) f(a + n ). n 1 k=0 f(a + (k+1)(b a) n ). n, a + If te upper sum and lower sum converges to a common limit S as n, S must be te area. Example: find te area below f(x) = x, x [0, 1]. Example: (Arcimedes, c. 250 BCE) find te area of a unit disc from te area of regular n-gons. Example: find te area below f(x) = sin x, x [0, π/2]. Solution: Te n-t lower sum is π n 1 kπ 2n k=0 sin 2n = π 1 2n sin π sin π n 1 kπ 4n k=0 sin 2n = π 2n sin π 4n 4n n 1 k=0 [cos( (k 1 2 )π 2n ) cos( (k+ 1 2 )π 2n )] = π 2n sin π [ n 1 k=0 cos( (k 1 2 )π 2n ) n k=1 cos( (k 1 2 )π 2n )] = π 2n sin π 4n 4n [cos( π 4n ) cos( π 2 π 4n )]. Wen n, te lower sum converges to 1. A similar calculation will sow tat te upper sum converges to 1 wen n. Hence, tis area is 1. Example: calculate te following: 10 k=1 2k Te summation symbol Let S = , ten S = Add te two equations, we get 2S = (1 + 19) + (3 + 17) + + (19 + 1) = = 200, so S = 100.
28 28 n k=0 ( 1)n Wen n is even, te sum is = 1 + ( 1 + 1) + ( 1 + 1) + ( 1 + 1) + + ( 1 + 1) = 1. Wen n is odd, te sum is = (1 1) + (1 1) + + (1 1) = k=0 3 n Let S = 100 k=0 3 n, ten 1 3 S = k=0 3 n = 101 k=1 3 n. Hence, 1 3 S S = = , S = In general, wen p 1, N n=0 pn = pn+1 1 p Riemann integration Definition 1. Partition finite interval [a, b] into n subintervals wit end points a = x 0 < x 1 < < x n = b, coose c k [x k 1, x k ], ten n k=1 f(c k)(x k x k 1 ) is called te Riemann sum. If, as te partition gets finer, for arbitrary coice of c k, te Riemann sum converges to a number I, ten we call I te Riemann integral of f from a to b, denoted as b a f(x)dx. Note tat tis definition works only for bounded function on finite interval. Wen b < a we define b a fdx = a b fdx. As a (somewat callenging) exercise on te ɛ δ language, one can prove te following: Teorem 9. b a f(x)dx exists if and only if te upper sum and lower sum defined in section 5.1 bot converges to te same limit as n. And b a f(x)dx equals tat limit. Geometrically, te definite integral is te signed area between te grap of f and te x-axis. Properties: (1) b a c b 1f(x) + c 2 g(x)dx = c 1 a f(x)dx + c b 2 a g(x)dx. (2) b a f(x)dx + c b f(x)dx = c a f(x)dx. (3) f 0, a < b, ten b a Example: Calculate: (1) b a 1dx (2) 1 1 f(x)dx 0. f(x)dx suc tat f(0) = 0 and f(x) = 1 for x 0 (3) 1 1 x dx (4) x 2 dx
29 (5) x 2 x dx (6) 3 0 f(x)dx were f(x) = 1 x wen 0 x 2 and f(x) = x 3 wen 2 < x Fundamental Teorem of Calculus Teorem 10. If f is continuous on [a, b], ten (1) d x dt a f(t)dt = f(x) (2) If F = f, ten F (b) F (a) = b a f(x)dx Examples: Calculate: (1) x (2) f(x)dx, were f(x) = d x 2 (3) dx 0 e t2 dt (4) x + ex dx { sin x, x 0 x, x > 0. Remark 7. b a f(t)dt and f(t)dt look similar but tey are conceptually COMPLETELY DIFFERENT. Fixing f, te former formula depends on a and b, wile te second formula is a function of t. Te FTC tells us tat wen f is continuous, tey are related by te following formula: f(x)dx = x a f(t)dt + C. Teorem 11. Any continuous function is integrable. Teorem 12. (Intrgral MVT) If f is continuous, tere is some c in [a, b] suc tat f(c) = b a f(t)dt Tis is a consequence of te intermediate value teorem and positivity. Tis is weaker tan te MVT but sometimes more convenient. Substitution From Cain rule F (g(x)) = g (x)f (g(x)), we know tat if te function we are integrating can be written in te form of g (F (g(x))) ten its indefinite integral is F (g(x)). Or, u (x)f(u(x))dx = f(u)du. 29 Example: sin x cos x 2 dx sin(2x 1)dx
30 30 Review for prelim Concavity and sketcing Concave up: f increase. Concave down: f decrease. Inflection point: f exist or is infinity, and concavity canges. Second derivative test: f (c) = 0, f (c) < 0 implies local max, f (c) = 0, f (c) > 0 implies local min. Graping: domain, symmetry, critical points, inflection points, increasing/decreasing, concave up/concave down, asymptotes, intersection wit x- and y-axis. Example: Sketc f(x) = x 3 e x. 4.5 L Hospital s Rule To use L Hospital s rule, we must make sure (1) te limit is of te form of 0 0 or, and (2) te derivative of te numerator and denominator as a limit. Example: lim x 0 (sin x) tan x 4.6 Example: Te longest side of a rigt-angled triangle as lengt 1. Wat is its largest possible area? 4.8 Indefinite integral Remember te constant C wen calculating indefinite integral. Capter 5: Example: Find te area between y = x 3 x and te x axis. Calculate 2 dx 1 5 2x. 5.6 Substitution for definite integral, symmetry, area between curves Substitution for definite integral: b a u (x)f(u(x))dx = u(b) u(a) f(u)du. Example: π/3 π/4 (cos x)1/3 sin 3 xdx. Symmetry: Example: 1 sin x 1 x dx. Area between curves: (1) Find te intersection points first. Example: find te area between y = sin x, y = cos x, x = 0 and x = 2π. (2) Sometimes it is more convenient to integrate in te y direction. Example: 1 0 sin 1 xdx. More examples of substitution: sec xdx (Hint: let u = sin x)
31 dx a (Hint: let x = a sin t) 2 x2 31
Exam 1 Review Solutions
Exam Review Solutions Please also review te old quizzes, and be sure tat you understand te omework problems. General notes: () Always give an algebraic reason for your answer (graps are not sufficient),
More information1 Calculus. 1.1 Gradients and the Derivative. Q f(x+h) f(x)
Calculus. Gradients and te Derivative Q f(x+) δy P T δx R f(x) 0 x x+ Let P (x, f(x)) and Q(x+, f(x+)) denote two points on te curve of te function y = f(x) and let R denote te point of intersection of
More informationA.P. CALCULUS (AB) Outline Chapter 3 (Derivatives)
A.P. CALCULUS (AB) Outline Capter 3 (Derivatives) NAME Date Previously in Capter 2 we determined te slope of a tangent line to a curve at a point as te limit of te slopes of secant lines using tat point
More information(a) At what number x = a does f have a removable discontinuity? What value f(a) should be assigned to f at x = a in order to make f continuous at a?
Solutions to Test 1 Fall 016 1pt 1. Te grap of a function f(x) is sown at rigt below. Part I. State te value of eac limit. If a limit is infinite, state weter it is or. If a limit does not exist (but is
More information1. Questions (a) through (e) refer to the graph of the function f given below. (A) 0 (B) 1 (C) 2 (D) 4 (E) does not exist
Mat 1120 Calculus Test 2. October 18, 2001 Your name Te multiple coice problems count 4 points eac. In te multiple coice section, circle te correct coice (or coices). You must sow your work on te oter
More informationSection 2.7 Derivatives and Rates of Change Part II Section 2.8 The Derivative as a Function. at the point a, to be. = at time t = a is
Mat 180 www.timetodare.com Section.7 Derivatives and Rates of Cange Part II Section.8 Te Derivative as a Function Derivatives ( ) In te previous section we defined te slope of te tangent to a curve wit
More informationMAT 145. Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points
MAT 15 Test #2 Name Solution Guide Type of Calculator Used TI-89 Titanium 100 points Score 100 possible points Use te grap of a function sown ere as you respond to questions 1 to 8. 1. lim f (x) 0 2. lim
More informationChapter 2 Limits and Continuity
4 Section. Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 6) Quick Review.. f () ( ) () 4 0. f () 4( ) 4. f () sin sin 0 4. f (). 4 4 4 6. c c c 7. 8. c d d c d d c d c 9. 8 ( )(
More informationChapter 2. Limits and Continuity 16( ) 16( 9) = = 001. Section 2.1 Rates of Change and Limits (pp ) Quick Review 2.1
Capter Limits and Continuity Section. Rates of Cange and Limits (pp. 969) Quick Review..... f ( ) ( ) ( ) 0 ( ) f ( ) f ( ) sin π sin π 0 f ( ). < < < 6. < c c < < c 7. < < < < < 8. 9. 0. c < d d < c
More information1. Consider the trigonometric function f(t) whose graph is shown below. Write down a possible formula for f(t).
. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd, periodic function tat as been sifted upwards, so we will use
More informationHOMEWORK HELP 2 FOR MATH 151
HOMEWORK HELP 2 FOR MATH 151 Here we go; te second round of omework elp. If tere are oters you would like to see, let me know! 2.4, 43 and 44 At wat points are te functions f(x) and g(x) = xf(x)continuous,
More informationClick here to see an animation of the derivative
Differentiation Massoud Malek Derivative Te concept of derivative is at te core of Calculus; It is a very powerful tool for understanding te beavior of matematical functions. It allows us to optimize functions,
More information1 The concept of limits (p.217 p.229, p.242 p.249, p.255 p.256) 1.1 Limits Consider the function determined by the formula 3. x since at this point
MA00 Capter 6 Calculus and Basic Linear Algebra I Limits, Continuity and Differentiability Te concept of its (p.7 p.9, p.4 p.49, p.55 p.56). Limits Consider te function determined by te formula f Note
More informationKey Concepts. Important Techniques. 1. Average rate of change slope of a secant line. You will need two points ( a, the formula: to find value
AB Calculus Unit Review Key Concepts Average and Instantaneous Speed Definition of Limit Properties of Limits One-sided and Two-sided Limits Sandwic Teorem Limits as x ± End Beaviour Models Continuity
More informationSome Review Problems for First Midterm Mathematics 1300, Calculus 1
Some Review Problems for First Midterm Matematics 00, Calculus. Consider te trigonometric function f(t) wose grap is sown below. Write down a possible formula for f(t). Tis function appears to be an odd,
More informationALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU. A. Fundamental identities Throughout this section, a and b denotes arbitrary real numbers.
ALGEBRA AND TRIGONOMETRY REVIEW by Dr TEBOU, FIU A. Fundamental identities Trougout tis section, a and b denotes arbitrary real numbers. i) Square of a sum: (a+b) =a +ab+b ii) Square of a difference: (a-b)
More informationDifferentiation Rules and Formulas
Differentiation Rules an Formulas Professor D. Olles December 1, 01 1 Te Definition of te Derivative Consier a function y = f(x) tat is continuous on te interval a, b]. Ten, te slope of te secant line
More informationName: Answer Key No calculators. Show your work! 1. (21 points) All answers should either be,, a (finite) real number, or DNE ( does not exist ).
Mat - Final Exam August 3 rd, Name: Answer Key No calculators. Sow your work!. points) All answers sould eiter be,, a finite) real number, or DNE does not exist ). a) Use te grap of te function to evaluate
More informationDerivatives. By: OpenStaxCollege
By: OpenStaxCollege Te average teen in te United States opens a refrigerator door an estimated 25 times per day. Supposedly, tis average is up from 10 years ago wen te average teenager opened a refrigerator
More informationMVT and Rolle s Theorem
AP Calculus CHAPTER 4 WORKSHEET APPLICATIONS OF DIFFERENTIATION MVT and Rolle s Teorem Name Seat # Date UNLESS INDICATED, DO NOT USE YOUR CALCULATOR FOR ANY OF THESE QUESTIONS In problems 1 and, state
More informationDerivatives. if such a limit exists. In this case when such a limit exists, we say that the function f is differentiable.
Derivatives 3. Derivatives Definition 3. Let f be a function an a < b be numbers. Te average rate of cange of f from a to b is f(b) f(a). b a Remark 3. Te average rate of cange of a function f from a to
More informationContinuity and Differentiability Worksheet
Continuity and Differentiability Workseet (Be sure tat you can also do te grapical eercises from te tet- Tese were not included below! Typical problems are like problems -3, p. 6; -3, p. 7; 33-34, p. 7;
More informationSolution. Solution. f (x) = (cos x)2 cos(2x) 2 sin(2x) 2 cos x ( sin x) (cos x) 4. f (π/4) = ( 2/2) ( 2/2) ( 2/2) ( 2/2) 4.
December 09, 20 Calculus PracticeTest s Name: (4 points) Find te absolute extrema of f(x) = x 3 0 on te interval [0, 4] Te derivative of f(x) is f (x) = 3x 2, wic is zero only at x = 0 Tus we only need
More informationSECTION 3.2: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY
(Section 3.2: Derivative Functions and Differentiability) 3.2.1 SECTION 3.2: DERIVATIVE FUNCTIONS and DIFFERENTIABILITY LEARNING OBJECTIVES Know, understand, and apply te Limit Definition of te Derivative
More information2.1 THE DEFINITION OF DERIVATIVE
2.1 Te Derivative Contemporary Calculus 2.1 THE DEFINITION OF DERIVATIVE 1 Te grapical idea of a slope of a tangent line is very useful, but for some uses we need a more algebraic definition of te derivative
More information2.3 More Differentiation Patterns
144 te derivative 2.3 More Differentiation Patterns Polynomials are very useful, but tey are not te only functions we need. Tis section uses te ideas of te two previous sections to develop tecniques for
More information. Compute the following limits.
Today: Tangent Lines and te Derivative at a Point Warmup:. Let f(x) =x. Compute te following limits. f( + ) f() (a) lim f( +) f( ) (b) lim. Let g(x) = x. Compute te following limits. g(3 + ) g(3) (a) lim
More informationPre-Calculus Review Preemptive Strike
Pre-Calculus Review Preemptive Strike Attaced are some notes and one assignment wit tree parts. Tese are due on te day tat we start te pre-calculus review. I strongly suggest reading troug te notes torougly
More informationGradient Descent etc.
1 Gradient Descent etc EE 13: Networked estimation and control Prof Kan) I DERIVATIVE Consider f : R R x fx) Te derivative is defined as d fx) = lim dx fx + ) fx) Te cain rule states tat if d d f gx) )
More informationTest 2 Review. 1. Find the determinant of the matrix below using (a) cofactor expansion and (b) row reduction. A = 3 2 =
Test Review Find te determinant of te matrix below using (a cofactor expansion and (b row reduction Answer: (a det + = (b Observe R R R R R R R R R Ten det B = (((det Hence det Use Cramer s rule to solve:
More informationSECTION 1.10: DIFFERENCE QUOTIENTS LEARNING OBJECTIVES
(Section.0: Difference Quotients).0. SECTION.0: DIFFERENCE QUOTIENTS LEARNING OBJECTIVES Define average rate of cange (and average velocity) algebraically and grapically. Be able to identify, construct,
More informationFinal Exam Review Exercise Set A, Math 1551, Fall 2017
Final Exam Review Exercise Set A, Math 1551, Fall 2017 This review set gives a list of topics that we explored throughout this course, as well as a few practice problems at the end of the document. A complete
More informationf a h f a h h lim lim
Te Derivative Te derivative of a function f at a (denoted f a) is f a if tis it exists. An alternative way of defining f a is f a x a fa fa fx fa x a Note tat te tangent line to te grap of f at te point
More information1 2 x Solution. The function f x is only defined when x 0, so we will assume that x 0 for the remainder of the solution. f x. f x h f x.
Problem. Let f x x. Using te definition of te derivative prove tat f x x Solution. Te function f x is only defined wen x 0, so we will assume tat x 0 for te remainder of te solution. By te definition of
More informationCalculus I - Spring 2014
NAME: Calculus I - Spring 04 Midterm Exam I, Marc 5, 04 In all non-multiple coice problems you are required to sow all your work and provide te necessary explanations everywere to get full credit. In all
More informationThe Derivative as a Function
Section 2.2 Te Derivative as a Function 200 Kiryl Tsiscanka Te Derivative as a Function DEFINITION: Te derivative of a function f at a number a, denoted by f (a), is if tis limit exists. f (a) f(a + )
More informationIntroduction to Derivatives
Introduction to Derivatives 5-Minute Review: Instantaneous Rates and Tangent Slope Recall te analogy tat we developed earlier First we saw tat te secant slope of te line troug te two points (a, f (a))
More informationFoundations of Calculus. November 18, 2014
Foundations of Calculus November 18, 2014 Contents 1 Conic Sections 3 11 A review of the coordinate system 3 12 Conic Sections 4 121 Circle 4 122 Parabola 5 123 Ellipse 5 124 Hyperbola 6 2 Review of Functions
More informationNotes: DERIVATIVES. Velocity and Other Rates of Change
Notes: DERIVATIVES Velocity and Oter Rates of Cange I. Average Rate of Cange A.) Def.- Te average rate of cange of f(x) on te interval [a, b] is f( b) f( a) b a secant ( ) ( ) m troug a, f ( a ) and b,
More informationDepartment of Mathematics, K.T.H.M. College, Nashik F.Y.B.Sc. Calculus Practical (Academic Year )
F.Y.B.Sc. Calculus Practical (Academic Year 06-7) Practical : Graps of Elementary Functions. a) Grap of y = f(x) mirror image of Grap of y = f(x) about X axis b) Grap of y = f( x) mirror image of Grap
More informationBob Brown Math 251 Calculus 1 Chapter 3, Section 1 Completed 1 CCBC Dundalk
Bob Brown Mat 251 Calculus 1 Capter 3, Section 1 Completed 1 Te Tangent Line Problem Te idea of a tangent line first arises in geometry in te context of a circle. But before we jump into a discussion of
More informationTime (hours) Morphine sulfate (mg)
Mat Xa Fall 2002 Review Notes Limits and Definition of Derivative Important Information: 1 According to te most recent information from te Registrar, te Xa final exam will be eld from 9:15 am to 12:15
More informationSolutions Manual for Precalculus An Investigation of Functions
Solutions Manual for Precalculus An Investigation of Functions David Lippman, Melonie Rasmussen 1 st Edition Solutions created at Te Evergreen State College and Soreline Community College 1.1 Solutions
More informationCalculus I Homework: The Derivative as a Function Page 1
Calculus I Homework: Te Derivative as a Function Page 1 Example (2.9.16) Make a careful sketc of te grap of f(x) = sin x and below it sketc te grap of f (x). Try to guess te formula of f (x) from its grap.
More informationMath 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006
Mat 102 TEST CHAPTERS 3 & 4 Solutions & Comments Fall 2006 f(x+) f(x) 10 1. For f(x) = x 2 + 2x 5, find ))))))))) and simplify completely. NOTE: **f(x+) is NOT f(x)+! f(x+) f(x) (x+) 2 + 2(x+) 5 ( x 2
More informationSection 15.6 Directional Derivatives and the Gradient Vector
Section 15.6 Directional Derivatives and te Gradient Vector Finding rates of cange in different directions Recall tat wen we first started considering derivatives of functions of more tan one variable,
More information2.8 The Derivative as a Function
.8 Te Derivative as a Function Typically, we can find te derivative of a function f at many points of its domain: Definition. Suppose tat f is a function wic is differentiable at every point of an open
More informationMAT 1339-S14 Class 2
MAT 1339-S14 Class 2 July 07, 2014 Contents 1 Rate of Cange 1 1.5 Introduction to Derivatives....................... 1 2 Derivatives 5 2.1 Derivative of Polynomial function.................... 5 2.2 Te
More information2.11 That s So Derivative
2.11 Tat s So Derivative Introduction to Differential Calculus Just as one defines instantaneous velocity in terms of average velocity, we now define te instantaneous rate of cange of a function at a point
More information2.3 Algebraic approach to limits
CHAPTER 2. LIMITS 32 2.3 Algebraic approac to its Now we start to learn ow to find its algebraically. Tis starts wit te simplest possible its, and ten builds tese up to more complicated examples. Fact.
More informationSin, Cos and All That
Sin, Cos and All Tat James K. Peterson Department of Biological Sciences and Department of Matematical Sciences Clemson University Marc 9, 2017 Outline Sin, Cos and all tat! A New Power Rule Derivatives
More informationLearning Objectives for Math 165
Learning Objectives for Math 165 Chapter 2 Limits Section 2.1: Average Rate of Change. State the definition of average rate of change Describe what the rate of change does and does not tell us in a given
More informationHow to Find the Derivative of a Function: Calculus 1
Introduction How to Find te Derivative of a Function: Calculus 1 Calculus is not an easy matematics course Te fact tat you ave enrolled in suc a difficult subject indicates tat you are interested in te
More information= 0 and states ''hence there is a stationary point'' All aspects of the proof dx must be correct (c)
Paper 1: Pure Matematics 1 Mark Sceme 1(a) (i) (ii) d d y 3 1x 4x x M1 A1 d y dx 1.1b 1.1b 36x 48x A1ft 1.1b Substitutes x = into teir dx (3) 3 1 4 Sows d y 0 and states ''ence tere is a stationary point''
More information4. The slope of the line 2x 7y = 8 is (a) 2/7 (b) 7/2 (c) 2 (d) 2/7 (e) None of these.
Mat 11. Test Form N Fall 016 Name. Instructions. Te first eleven problems are wort points eac. Te last six problems are wort 5 points eac. For te last six problems, you must use relevant metods of algebra
More information3.4 Worksheet: Proof of the Chain Rule NAME
Mat 1170 3.4 Workseet: Proof of te Cain Rule NAME Te Cain Rule So far we are able to differentiate all types of functions. For example: polynomials, rational, root, and trigonometric functions. We are
More informationHigher Derivatives. Differentiable Functions
Calculus 1 Lia Vas Higer Derivatives. Differentiable Functions Te second derivative. Te derivative itself can be considered as a function. Te instantaneous rate of cange of tis function is te second derivative.
More informationContinuity. Example 1
Continuity MATH 1003 Calculus and Linear Algebra (Lecture 13.5) Maoseng Xiong Department of Matematics, HKUST A function f : (a, b) R is continuous at a point c (a, b) if 1. x c f (x) exists, 2. f (c)
More informationSFU UBC UNBC Uvic Calculus Challenge Examination June 5, 2008, 12:00 15:00
SFU UBC UNBC Uvic Calculus Callenge Eamination June 5, 008, :00 5:00 Host: SIMON FRASER UNIVERSITY First Name: Last Name: Scool: Student signature INSTRUCTIONS Sow all your work Full marks are given only
More informationSection 2.1 The Definition of the Derivative. We are interested in finding the slope of the tangent line at a specific point.
Popper 6: Review of skills: Find tis difference quotient. f ( x ) f ( x) if f ( x) x Answer coices given in audio on te video. Section.1 Te Definition of te Derivative We are interested in finding te slope
More informationMA119-A Applied Calculus for Business Fall Homework 4 Solutions Due 9/29/ :30AM
MA9-A Applied Calculus for Business 006 Fall Homework Solutions Due 9/9/006 0:0AM. #0 Find te it 5 0 + +.. #8 Find te it. #6 Find te it 5 0 + + = (0) 5 0 (0) + (0) + =.!! r + +. r s r + + = () + 0 () +
More informationLecture XVII. Abstract We introduce the concept of directional derivative of a scalar function and discuss its relation with the gradient operator.
Lecture XVII Abstract We introduce te concept of directional derivative of a scalar function and discuss its relation wit te gradient operator. Directional derivative and gradient Te directional derivative
More informationCHAPTER 3: Derivatives
CHAPTER 3: Derivatives 3.1: Derivatives, Tangent Lines, and Rates of Cange 3.2: Derivative Functions and Differentiability 3.3: Tecniques of Differentiation 3.4: Derivatives of Trigonometric Functions
More informationDRAFT - Math 101 Lecture Note - Dr. Said Algarni
3 Differentiation Rules 3.1 The Derivative of Polynomial and Exponential Functions In this section we learn how to differentiate constant functions, power functions, polynomials, and exponential functions.
More informationCalculus I Practice Exam 1A
Calculus I Practice Exam A Calculus I Practice Exam A Tis practice exam empasizes conceptual connections and understanding to a greater degree tan te exams tat are usually administered in introductory
More informationRecall from our discussion of continuity in lecture a function is continuous at a point x = a if and only if
Computational Aspects of its. Keeping te simple simple. Recall by elementary functions we mean :Polynomials (including linear and quadratic equations) Eponentials Logaritms Trig Functions Rational Functions
More informationExercises for numerical differentiation. Øyvind Ryan
Exercises for numerical differentiation Øyvind Ryan February 25, 2013 1. Mark eac of te following statements as true or false. a. Wen we use te approximation f (a) (f (a +) f (a))/ on a computer, we can
More informationCombining functions: algebraic methods
Combining functions: algebraic metods Functions can be added, subtracted, multiplied, divided, and raised to a power, just like numbers or algebra expressions. If f(x) = x 2 and g(x) = x + 2, clearly f(x)
More informationDifferential Calculus Definitions, Rules and Theorems
Precalculus Review Differential Calculus Definitions, Rules an Teorems Sara Brewer, Alabama Scool of Mat an Science Functions, Domain an Range f: X Y a function f from X to Y assigns to eac x X a unique
More information. If lim. x 2 x 1. f(x+h) f(x)
Review of Differential Calculus Wen te value of one variable y is uniquely determined by te value of anoter variable x, ten te relationsip between x and y is described by a function f tat assigns a value
More informationMaterial for Difference Quotient
Material for Difference Quotient Prepared by Stepanie Quintal, graduate student and Marvin Stick, professor Dept. of Matematical Sciences, UMass Lowell Summer 05 Preface Te following difference quotient
More informationMathematics 123.3: Solutions to Lab Assignment #5
Matematics 3.3: Solutions to Lab Assignment #5 Find te derivative of te given function using te definition of derivative. State te domain of te function and te domain of its derivative..: f(x) 6 x Solution:
More informationCHAPTER 3: DERIVATIVES
(Answers to Exercises for Capter 3: Derivatives) A.3.1 CHAPTER 3: DERIVATIVES SECTION 3.1: DERIVATIVES, TANGENT LINES, and RATES OF CHANGE 1) a) f ( 3) f 3.1 3.1 3 f 3.01 f ( 3) 3.01 3 f 3.001 f ( 3) 3.001
More informationMath Spring 2013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, (1/z) 2 (1/z 1) 2 = lim
Mat 311 - Spring 013 Solutions to Assignment # 3 Completion Date: Wednesday May 15, 013 Question 1. [p 56, #10 (a)] 4z Use te teorem of Sec. 17 to sow tat z (z 1) = 4. We ave z 4z (z 1) = z 0 4 (1/z) (1/z
More informationMath 1210 Midterm 1 January 31st, 2014
Mat 110 Midterm 1 January 1st, 01 Tis exam consists of sections, A and B. Section A is conceptual, wereas section B is more computational. Te value of every question is indicated at te beginning of it.
More informationREVIEW LAB ANSWER KEY
REVIEW LAB ANSWER KEY. Witout using SN, find te derivative of eac of te following (you do not need to simplify your answers): a. f x 3x 3 5x x 6 f x 3 3x 5 x 0 b. g x 4 x x x notice te trick ere! x x g
More information5.1 We will begin this section with the definition of a rational expression. We
Basic Properties and Reducing to Lowest Terms 5.1 We will begin tis section wit te definition of a rational epression. We will ten state te two basic properties associated wit rational epressions and go
More informationContinuity and Differentiability of the Trigonometric Functions
[Te basis for te following work will be te definition of te trigonometric functions as ratios of te sides of a triangle inscribed in a circle; in particular, te sine of an angle will be defined to be te
More informationDifferentiation in higher dimensions
Capter 2 Differentiation in iger dimensions 2.1 Te Total Derivative Recall tat if f : R R is a 1-variable function, and a R, we say tat f is differentiable at x = a if and only if te ratio f(a+) f(a) tends
More informationMATH 1A Midterm Practice September 29, 2014
MATH A Midterm Practice September 9, 04 Name: Problem. (True/False) If a function f : R R is injective, ten f as an inverse. Solution: True. If f is injective, ten it as an inverse since tere does not
More informationCalculus AB Topics Limits Continuity, Asymptotes
Calculus AB Topics Limits Continuity, Asymptotes Consider f x 2x 1 x 3 1 x 3 x 3 Is there a vertical asymptote at x = 3? Do not give a Precalculus answer on a Calculus exam. Consider f x 2x 1 x 3 1 x 3
More informationSECTION 2.1 BASIC CALCULUS REVIEW
Tis capter covers just te very basics of wat you will nee moving forwar onto te subsequent capters. Tis is a summary capter, an will not cover te concepts in-ept. If you ve never seen calculus before,
More informationThe derivative function
Roberto s Notes on Differential Calculus Capter : Definition of derivative Section Te derivative function Wat you need to know already: f is at a point on its grap and ow to compute it. Wat te derivative
More information1 (10) 2 (10) 3 (10) 4 (10) 5 (10) 6 (10) Total (60)
First Name: OSU Number: Last Name: Signature: OKLAHOMA STATE UNIVERSITY Department of Matematics MATH 2144 (Calculus I) Instructor: Dr. Matias Sculze MIDTERM 1 September 17, 2008 Duration: 50 minutes No
More informationa x a y = a x+y a x a = y ax y (a x ) r = a rx and log a (xy) = log a (x) + log a (y) log a ( x y ) = log a(x) log a (y) log a (x r ) = r log a (x).
You should prepare the following topics for our final exam. () Pre-calculus. (2) Inverses. (3) Algebra of Limits. (4) Derivative Formulas and Rules. (5) Graphing Techniques. (6) Optimization (Maxima and
More informationThe Derivative The rate of change
Calculus Lia Vas Te Derivative Te rate of cange Knowing and understanding te concept of derivative will enable you to answer te following questions. Let us consider a quantity wose size is described by
More informationDerivatives of Exponentials
mat 0 more on derivatives: day 0 Derivatives of Eponentials Recall tat DEFINITION... An eponential function as te form f () =a, were te base is a real number a > 0. Te domain of an eponential function
More informationLesson 6: The Derivative
Lesson 6: Te Derivative Def. A difference quotient for a function as te form f(x + ) f(x) (x + ) x f(x + x) f(x) (x + x) x f(a + ) f(a) (a + ) a Notice tat a difference quotient always as te form of cange
More information(e) 2 (f) 2. (c) + (d). Limits at Infinity. 2.5) 9-14,25-34,41-43,46-47,56-57, (c) (d) 2
Math 150A. Final Review Answers, Spring 2018. Limits. 2.2) 7-10, 21-24, 28-1, 6-8, 4-44. 1. Find the values, or state they do not exist. (a) (b) 1 (c) DNE (d) 1 (e) 2 (f) 2 (g) 2 (h) 4 2. lim f(x) = 2,
More informationSection 3: The Derivative Definition of the Derivative
Capter 2 Te Derivative Business Calculus 85 Section 3: Te Derivative Definition of te Derivative Returning to te tangent slope problem from te first section, let's look at te problem of finding te slope
More informationChapter 2: Functions, Limits and Continuity
Chapter 2: Functions, Limits and Continuity Functions Limits Continuity Chapter 2: Functions, Limits and Continuity 1 Functions Functions are the major tools for describing the real world in mathematical
More informationSolutions to Math 41 Final Exam December 10, 2012
Solutions to Math 4 Final Exam December,. ( points) Find each of the following limits, with justification. If there is an infinite limit, then explain whether it is or. x ln(t + ) dt (a) lim x x (5 points)
More information. h I B. Average velocity can be interpreted as the slope of a tangent line. I C. The difference quotient program finds the exact value of f ( a)
Capter Review Packet (questions - ) KEY. In eac case determine if te information or statement is correct (C) or incorrect (I). If it is incorrect, include te correction. f ( a ) f ( a) I A. represents
More informationChapter 2 Limits and Continuity. Section 2.1 Rates of Change and Limits (pp ) Section Quick Review 2.1
Section. 6. (a) N(t) t (b) days: 6 guppies week: 7 guppies (c) Nt () t t t ln ln t ln ln ln t 8. 968 Tere will be guppies ater ln 8.968 days, or ater nearly 9 days. (d) Because it suggests te number o
More informationlim 1 lim 4 Precalculus Notes: Unit 10 Concepts of Calculus
Syllabus Objectives: 1.1 Te student will understand and apply te concept of te limit of a function at given values of te domain. 1. Te student will find te limit of a function at given values of te domain.
More informationAnalytic Functions. Differentiable Functions of a Complex Variable
Analytic Functions Differentiable Functions of a Complex Variable In tis capter, we sall generalize te ideas for polynomials power series of a complex variable we developed in te previous capter to general
More informationMath 212-Lecture 9. For a single-variable function z = f(x), the derivative is f (x) = lim h 0
3.4: Partial Derivatives Definition Mat 22-Lecture 9 For a single-variable function z = f(x), te derivative is f (x) = lim 0 f(x+) f(x). For a function z = f(x, y) of two variables, to define te derivatives,
More informationNumerical Differentiation
Numerical Differentiation Finite Difference Formulas for te first derivative (Using Taylor Expansion tecnique) (section 8.3.) Suppose tat f() = g() is a function of te variable, and tat as 0 te function
More informationCalculus I, Fall Solutions to Review Problems II
Calculus I, Fall 202 - Solutions to Review Problems II. Find te following limits. tan a. lim 0. sin 2 b. lim 0 sin 3. tan( + π/4) c. lim 0. cos 2 d. lim 0. a. From tan = sin, we ave cos tan = sin cos =
More informationTopics and Concepts. 1. Limits
Topics and Concepts 1. Limits (a) Evaluating its (Know: it exists if and only if the it from the left is the same as the it from the right) (b) Infinite its (give rise to vertical asymptotes) (c) Limits
More information