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 0 2.2 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 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 2 1 + 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.
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 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 2.718 2 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 10001 + 56x 345 + 1000005x 2 + x + 345680907 as a real root. (Hint: use Intermediate Value Teorem.)
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 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 { b, x = 0 4. Find b suc tat f(x) = e 1 is continuous. x 2, x 0
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
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 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 2 +1. Additional Example: f(x) = { 0, x = 0 x 3 sin 1 x, at x = 0. (int: use sandwic teorem) x 0 3.3 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 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 2 2 0+ 0 = 0, lim 0 + = 2 lim 0 +( sin cos 1 cos ) = lim 0 sin + lim 0 1 + cos lim 2 sin 2 0 + 2 lim ( 0 2 )2 + 4 = 0, so
12 sin lim 0 + 2 g(x) = (x sin x ) = 0. q.e.d. { 0, x = 0 x sin 1 find g x, x 0,
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. 13 3.5-3.6, 3.8-3.9 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 (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).
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 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 ) 2 4 + 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.
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 1. 3. (tan 1 (ln(x))) Solution: By cain rule, tan 1 (ln(x))) = 1 x 1+(ln(x)) 2.. 1 1 x 2
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 3.1-3.7 3.1.18, 3.3.64: te slope of te tangent line is te derivative. 3.1.32: 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 = 0 3.2.8, 3.2.12, 3.3.2, 3.3.6, 3.3.36, 3.5.2, 3.5.6, 3.5.18, 3.5.26, 3.6.6, 3.6.18, 3.6.28, 3.6.34, 3.6.55, 3.6.82: calculation of derivative wit te rules. 3.3.54: use differentiation rules. 3.4.10: maximum iff derivative is 0.
19 (*)3.2.28, 3.2.30, 3.4.10: sketcing derivative. (*)3.5.57: use te limit of sin(x)/x. (*)3.7.16, 3.7.20, 3.7.30: 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 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. 3.11 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).
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. 21 4.1 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 6. 1. 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 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 1. 1. 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
Solution: x 4 x 2 = { x 4 x 2, x 1 or x 1 x 2 x 4, 1 < x < 1 23. 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) 2 2 2 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 1 2 + 2kπ, local minimum wen x = π 1 2 + 2kπ, 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. 4. 1 5 x5 2 3 x3 + x 1 17 Solution: Let f(x) = 1 5 x5 2 3 x3 + x 1 17. 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 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 7. 1. 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:
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. 25 4.6 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 2 + 2 1 r 2 + 2 2πr = πr(r + r 2 + 2 ). 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 derivative test we know tat V reaces its maximum wen r = 1 2 S is 3 2. 3 2 3 2 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 2 + 1 2 r 2 + 2 2πr = πr(r + r 2 + 2 ). Fix V, solve for wit respect to r we get = 3V, ence S = πr(r + r πr 2 + 2 ) = 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 3 2 3 2 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 5.1 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 1 5.2 Te summation symbol Let S = 1 + 3 + 5 + 7 + + 19, ten S = 19 + 17 + + 3 + 1. Add te two equations, we get 2S = (1 + 19) + (3 + 17) + + (19 + 1) = 20 10 = 200, so S = 100.
28 n k=0 ( 1)n Wen n is even, te sum is 1 1 + 1 1 + 1 1 + + 1 = 1 + ( 1 + 1) + ( 1 + 1) + ( 1 + 1) + + ( 1 + 1) = 1. Wen n is odd, te sum is 1 1 + 1 1 + 1 1 + + 1 1 = (1 1) + (1 1) + + (1 1) = 0. 100 k=0 3 n Let S = 100 k=0 3 n, ten 1 3 S = 1 3 100 k=0 3 n = 101 k=1 3 n. Hence, 1 3 S S = 3 101 3 0 = 3 101 1, S = 3 101 1 3 1 1. In general, wen p 1, N n=0 pn = pn+1 1 p 1. 5.3 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) 1 1 4 x 2 dx
(5) 1 1 4 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 3. 5.4 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) 2 1 1 x (2) f(x)dx, were f(x) = d x 2 (3) dx 0 e t2 dt (4) 1 1 1 + 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 Review for prelim 3 4.4 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)
dx a (Hint: let x = a sin t) 2 x2 31