Differentiation Rules Derivatives of Polynomials and Exponential Functions
|
|
- Paulina Carpenter
- 5 years ago
- Views:
Transcription
1 Derivatives of Polynomials an Exponential Functions Differentiation Rules Derivatives of Polynomials an Exponential Functions Let s start with the simplest of all functions, the constant function f(x) = c. The graph of this function is the horizontal line y = c, which has slope 0, so we must have f (x) = 0. A formal proof of a erivative, is also easy f f(x + h) f(x) c c (x) = lim = lim h 0 h h 0 h = lim 0 = 0. h 0 In Leibniz notation, we write this rule as follows. Derivative of a Constant Function (c) = 0. x MAT 1001 Calculus I 1 / 66
2 Derivatives of Polynomials an Exponential Functions Power Rules Power Rules We next look at the functions f(x) = x n, where n is a positive integer. If n = 1, the graph of f(x) = x is the line y = x, which has slope 1. So (x) = 1. x We can also verify that the cases n = 2, n = 3 an n = 4 have forms x (x2 ) = 2x, x (x3 ) = 3x 2, x (x4 ) = 4x 3. MAT 1001 Calculus I 2 / 66
3 Derivatives of Polynomials an Exponential Functions Power Rules Comparing the equations, we see a pattern emerging. It seems to be a reasonable guess that, when n is a positive integer, (/x)(x n ) = nx n 1. This turns out to be true. The Power Rule If n is a positive integer, then x (xn ) = nx n 1. Note The power rule is also vali for all real numbers n. MAT 1001 Calculus I 3 / 66
4 Derivatives of Polynomials an Exponential Functions New Derivatives from Ol New Derivatives from Ol The Constant Multiple Rule If c is a constant an f is a ifferentiable function, then x [cf(x)] = c x f(x). The Sum-Difference Rule If f an g are both ifferentiable, then [f(x) ± g(x)] = x x f(x) ± x g(x). MAT 1001 Calculus I 4 / 66
5 Derivatives of Polynomials an Exponential Functions Derivative of Exponential Functions Derivative of Exponential Functions Definition of the Number e e h 1 e is the number such that lim = 1. h 0 h Derivative of the Natural Exponential Function x (ex ) = e x Derivative of the Exponential Function For the real number a > 0, a 1 x (ax ) = a x ln a MAT 1001 Calculus I 5 / 66
6 The Prouct an Quotient Rules The Prouct an Quotient Rules The Prouct Rule If f an g are both ifferentiable, then [f(x)g(x)] = f(x) [g(x)] + g(x) x x x [f(x)]. The Quotient Rule If f an g are ifferentiable, then x [ ] f(x) g(x) [f(x)] f(x) = x x [g(x)] g(x) [g(x)] 2. MAT 1001 Calculus I 6 / 66
7 The Prouct an Quotient Rules The Secon Derivative The Secon Derivative If f is a ifferentiable function, then its erivative f is also a function, f so may have a erivative of its own, enote by (f ) = f. This new function f is calle the secon erivative of f because it is the erivative of the erivative of f. Using Leibniz notation, we write the secon erivative of as y = f(x) x ( ) y = 2 y x x 2. MAT 1001 Calculus I 7 / 66
8 The Prouct an Quotient Rules The Secon Derivative Example 1 Differentiate: (a) f(x) = 1 x 2 (b) y = 3 x 2 Solution. In each case we rewrite the function as a power of x. (a) Since f(x) = x 2, we use the Power Rule with n = 2: f (x) = x (x 2 ) = 2x 2 1 = 2x 3 = 2 x 3 (b) y x = x ( 3 x 2 ) = x (x2/3 ) = 2 3 x(2/3) 1 = 2 3 x 1/3 MAT 1001 Calculus I 8 / 66
9 The Prouct an Quotient Rules The Secon Derivative Example 2 x (x8 + 12x 5 4x x 3 6x + 5) = x (x8 ) + 12 x (x5 ) 4 x (x4 ) + 10 x (x3 ) 6 x (x) + x (5) = 8x (5x 4 ) 4(4x 3 ) + 10(3x 2 ) 6(1) + 0 = 8x x 4 16x x 2 6 MAT 1001 Calculus I 9 / 66
10 The Prouct an Quotient Rules The Secon Derivative Example 3 Fin the points on the curve y = x 4 6x where the tangent line is horizontal. Solution. Horizontal tangents occur where the erivative is zero. We have y x = x (x4 ) 6 x (x2 ) + x (4) = 4x 3 12x + 0 = 4x(x 2 3) Thus, y/x = 0 if x = 0 or x 2 3 = 0, that is, x = ± 3. So the given curve has horizontal tangents when x = 0, x = 3 an x = 3. The corresponing points are (0, 4), ( 3, 5) an ( 3, 5). MAT 1001 Calculus I 10 / 66
11 The Prouct an Quotient Rules 3, 5). (See Figure 5.) Solution (cont.) y The Secon Derivative (0, 4) 0 x {_œ 3, _5} {œ 3, _5} MAT 1001 Calculus I 11 / 66
12 The Prouct an Quotient Rules The Secon Derivative Example 4 Differentiate the function f(t) = t(1 t). Solution 1. Using the Prouct Rule, we have f (t) = t (1 t) + (1 t) x x ( t) = t( 1) + (1 t) 1 2 t 1/2 = t + 1 t 2 t = 1 3t 2 t MAT 1001 Calculus I 12 / 66
13 The Prouct an Quotient Rules The Secon Derivative Solution 2. If we first use the laws of exponents to rewrite f(t), then we can procee irectly without using the Prouct Rule. f(t) = t t t = t 1/2 t 3/2 f (t) = 1 2 t 1/2 3 2 t1/2 which is equivalent to the answer given in Solution 1. The previous example shows that it is sometimes easier to simplify a prouct of functions than to use the Prouct Rule. MAT 1001 Calculus I 13 / 66
14 The Prouct an Quotient Rules The Secon Derivative Example 5 If f(x) = x. g(x), where g(4) = 2 an g (4) = 3, fin f (4). Solution. Applying the Prouct Rule, we get f (x) = ( ) x. g(x) = x. x x (g(x)) + g(x). ( ) x x = x. g (x) + g(x) x 1/2 = x. g (x) + g(x) 2 x So f (4) = 4. g (4) + g(4) 2 4 = = 6.5 MAT 1001 Calculus I 14 / 66
15 The Prouct an Quotient Rules The Secon Derivative Example 6 Let y = x2 + x 2 x 3. Then + 6 y = (x 3 + 6) x (x2 + x 2) (x 2 + x 2) x (x3 + 6) (x 3 + 6) 2 = (x3 + 6)(2x + 1) (x 2 + x 2)(3x 2 ) (x 3 + 6) 2 = (2x4 + x x + 6) (3x 4 + 3x 3 6x 2 ) (x 3 + 6) 2 = x4 2x 3 + 6x x + 6 (x 3 + 6) 2. MAT 1001 Calculus I 15 / 66
16 The Prouct an Quotient Rules The Secon Derivative Note Although it is possible to ifferentiate the function F (x) = 3x2 + 2 x x using the Quotient Rule, it is much easier to perform the ivision first an write the function as F (x) = 3x + 2x 1/2 before ifferentiating. MAT 1001 Calculus I 16 / 66
17 The Prouct an Quotient Rules The Secon Derivative Example 7 If f(x) = e x x, fin f an f. Solution. Using the Difference Rule, we have f (x) = x (ex x) = x (ex ) x (x) = ex 1 In previous section we efine the secon erivative as the erivative of f, so f (x) = x (ex 1) = x (ex ) (1) = ex x MAT 1001 Calculus I 17 / 66
18 The Prouct an Quotient Rules The Secon Derivative Example 8 At what point on the curve y = e x is the tangent line parallel to the line y = 2x? Solution. Since y = e x, we have y = e x. Let the x-coorinate of the point in question be a. Then the slope of the tangent line at that point is e a. This tangent line will be parallel to the line y = 2x if it has the same slope, that is, 2. Equating slopes, we get e a = 2 a = ln 2 Therefore, the require point is (a, e a ) = (ln 2, 2). MAT 1001 Calculus I 18 / 66
19 The Prouct an Quotient Rules The Secon Derivative _2 2 0 Solution (cont.) FIGURE 8 y We know concave y= 3 (ln 2, 2) 2 1 y=2x EXAMPLE line y SOLUTION be a. Th parallel 0 1 x FIGURE 9 Therefor MAT 1001 Calculus I 19 / 66
20 The Prouct an Quotient Rules The Secon Derivative Example 9 (a) If f(x) = xe x, fin f (x). (b) Fin the nth erivative, f (n) (x). Solution. (a) By the Prouct Rule, we have f (x) = x (xex ) = x x (ex ) + e x x (x) = xe x + e x 1 = (x + 1)e x MAT 1001 Calculus I 20 / 66
21 The Prouct an Quotient Rules The Secon Derivative Solution (cont.) (b) Using the Prouct Rule a secon time, we get f (x) = x [(x + 1)ex ] = (x + 1) x (ex ) + e x (x + 1) x = (x + 1)e x + e x 1 = (x + 2)e x Further applications of the Prouct Rule give f (x) = (x + 3)e x f (4) (x) = (x + 4)e x In fact, each successive ifferentiation as another term, so f (n) (x) = (x + n)e x. MAT 1001 Calculus I 21 / 66
22 The Prouct an Quotient Rules The Secon Derivative Example 10 Fin an equation of the tangent line to the curve y = e x /(1 + x 2 ) at the point (1, e/2). Solution. Accoring to the Quotient Rule, we have y (1 + x2 ) x = x (ex ) e x x (1 + x2 ) (1 + x 2 ) 2 = (1 + x2 )e x e x (2x) (1 + x 2 ) 2 = ex (1 x) 2 (1 + x 2 ) 2 So the slope of the tangent line at (1, e/2) is y x = 0. x=1 MAT 1001 Calculus I 22 / 66
23 The Prouct an Quotient Rules The Secon Derivative Solution (cont.) This means that the tangent line at (1, e/2) is horizontal an its equation is y = e/2. [Notice that the function is increasing an crosses its tangent line at (1, e/2).] 2.5 y= 1+ So the slope _ FIGURE 4 This means [See Figure 1, e 2.] MAT 1001 Calculus I 23 / 66
24 Derivatives of Trigonometric Functions Derivatives of Trigonometric Functions (sin x) = cos x x (cos x) = sin x x x (tan x) = sec2 x (csc x) = csc x cot x x (sec x) = sec x tan x x x (cot x) = csc2 x MAT 1001 Calculus I 24 / 66
25 Derivatives of Trigonometric Functions Example 11 Differentiate y = x 2 sin x. Solution. Using the Prouct Rule, we have y x = x2 (sin x) + sin x x x (x2 ) = x 2 cos x + 2x sin x. MAT 1001 Calculus I 25 / 66
26 Derivatives of Trigonometric Functions Example 12 Differentiate f(x) = sec x. For what values of x oes the graph of f 1 + tan x have a horizontal tangent? Solution. The Quotient Rule gives f (x) = (1 + tan x) x (sec x) sec x x (1 + tan x) (1 + tan x) 2 = (1 + tan x) sec x tan x sec x sec2 x (1 + tan x) 2 = sec x [tan x + tan2 x sec 2 x] (1 + tan x) 2 MAT 1001 Calculus I 26 / 66
27 Derivatives of Trigonometric Functions Solution (cont.) f (x) = sec x (tan x 1) (1 + tan x) 2 In simplifying the answer we have use the ientity tan 2 x + 1 = sec 2 x. Since sec x is never 0, we see that f (x) = 0 when tan x = 1, an this occurs when x = nπ + π/4, where n is an integer. MAT 1001 Calculus I 27 / 66
28 Derivatives of Trigonometric Functions Example 13 Fin the 27th erivative of cos x. Solution. The first few erivatives f(x) = cos x are as follows: f (x) = sin x f (x) = cos x f (x) = sin x f (4) (x) = cos x f (5) (x) = sin x We see that the successive erivatives occur in a cycle of length 4 an, in particular, f (n) (x) = cos x whenever n is a multiple of 4. Therefore f (24) (x) = cos x an, ifferentiating three more times, we have f (27) (x) = sin x. MAT 1001 Calculus I 28 / 66
29 The Chain Rule The Chain Rule Suppose you are aske to ifferentiate the function F (x) = x The ifferentiation formulas you learne in the previous sections of this chapter o not enable you to calculate F (x). Observe that F is a composite function. In fact, if we let y = f(u) = u an let u = g(x) = x 2 + 1, then we can write y = F (x) = f(g(x)), that is, F = f g. MAT 1001 Calculus I 29 / 66
30 The Chain Rule Chain Rule If f an g are both ifferentiable an F = f g is the composite function efine by F (x) = f(g(x)), then F is ifferentiable an F is given by the prouct F (x) = f (g(x))g (x) (1) In Leibniz notation, if y = f(u) an u = g(x) are both ifferentiable functions, then y x = y u u x (2) MAT 1001 Calculus I 30 / 66
31 The Chain Rule Note In using the Chain Rule we work from the outsie to the insie. Formula (1) says that we ifferentiate the outer function f [at the inner function g(x)] an then we multiply by the erivative of the inner function. x f }{{} outer function (g(x)) }{{} evaluate at inner function = f }{{} erivative of outer function (g(x)) }{{} evaluate at inner function g (x) }{{} erivative of inner function MAT 1001 Calculus I 31 / 66
32 The Chain Rule Example 14 Fin F (x) if F (x) = x Solution 1. (using Equation (1)): At the beginning of this section we expresse F as F (x) = (f g)(x) = f(g(x)) where f(u) = u an g(x) = x Since we have f (u) = 1 2 u 1/2 = 1 2 u an g (x) = 2x, F (x) = f (g(x)) g (x) 1 = 2 x x = x x MAT 1001 Calculus I 32 / 66
33 The Chain Rule Solution 2. (using Equation (2)): If we let u = x an y = u, then F (x) = y u u x = 1 2 u 2x 1 = 2 x x = x x MAT 1001 Calculus I 33 / 66
34 The Chain Rule Example 15 Differentiate (a) y = sin(x 2 ) an (b) y = sin 2 x. Solution. (a) If y = sin(x 2 ), then the outer function is the sine function an the inner function is the squaring function, so the Chain Rule gives y x = x sin(x2 ) = cos(x 2 ) = 2x cos(x 2 ). x x2 MAT 1001 Calculus I 34 / 66
35 The Chain Rule Solution (cont.) (b) Note that sin 2 x = (sin x) 2. Here the outer function is the squaring function an the inner function is the sine function. So y x = x (sin x)2 = 2 sin x cos x The answer can be left as 2 sin x cos x or written as sin 2x (by a trigonometric ientity known as the ouble-angle formula). MAT 1001 Calculus I 35 / 66
36 The Chain Rule Example 16 Differentiate y = (x 3 1) 100. Solution. Using Chain Rule y x = x (x3 1) 100 = 100(x 3 1) 99 x (x3 1) = 100(x 3 1) 99 3x 2 = 300x 2 (x 3 1) 99. MAT 1001 Calculus I 36 / 66
37 The Chain Rule Example 17 Fin f (x) if f(x) = 1 3 x 2 + x + 1. Solution. First rewrite f as f(x) = (x 2 + x + 1) 1/3. Thus f (x) = 1 3 (x2 + x + 1) 4/3 x (x2 + x + 1) = 1 3 (x2 + x + 1) 4/3 (2x + 1). MAT 1001 Calculus I 37 / 66
38 The Chain Rule Example 18 Differentiate y = e sin x. Solution. Here the inner function is g(x) = sin x an the outer function is the exponential function f(x) = e x. So, by the Chain Rule, y x = x (esin x ) = e sin x x (sin x) = esin x cos x. MAT 1001 Calculus I 38 / 66
39 The Chain Rule Example 19 Differentiate: (a) f(x) = 1 x 2 (b) y = 3 x 2 Solution. Using Chain Rule, (a) f (x) = ( ) 1 x x 2 = 1 (x 2 ) 2 x (x2 ) = 1 (x 2 ) 2 (2x) = 2 x 3 (b) y x = x ( 3 x 2 ) = 1 3 x 2/3 x (x2 ) = 1 3 x 2/3 (2x) = 2 3 x 1/3 MAT 1001 Calculus I 39 / 66
40 The Chain Rule Example 20 If f(x) = sin(cos(tan x)), then f (x) = cos(cos(tan x)) cos(tan x) x = cos(cos(tan x))[ sin(tan x)] (tan x) x = cos(cos(tan x)) sin(tan x) sec 2 x. Notice that we use the Chain Rule twice. MAT 1001 Calculus I 40 / 66
41 The Chain Rule Example 21 Differentiate y = e sec 3θ. Solution. The outer function is the exponential function, the mile function is the secant function an the inner function is the tripling function. So, we have y θ = esec 3θ (sec 3θ) θ = e sec 3θ sec 3θ tan 3θ θ (3θ) = 3e sec 3θ sec 3θ tan 3θ. MAT 1001 Calculus I 41 / 66
42 Implicit Differentiation Implicit Differentiation The functions that we have met so far can be escribe by expressing one variable explicitly in terms of another variable-for example, y = x or y = x sin x or, in general, y = f(x). Some functions, however, are efine implicitly by a relation between x an y such as x 2 + y 2 = 25 (3) or x 3 + y 3 = 6xy. (4) MAT 1001 Calculus I 42 / 66
43 Implicit Differentiation In some cases it is possible to solve such an equation for y as an explicit function (or several functions) of x. For instance, if we solve Equation (3) for y, we obtain y = ± 25 x 2, RULES an so two functions etermine by the implicit Equation (3) are RULES ON RULES f(x) = x 2 an g(x) = f x f x f x s25 s25 x 2 xan 2 xan 2 an t x t x t x s25 s25 x 2 x. 2 The x. 2 The. The graphs graphs of of fof an f 25 an f an t txare t 2 the are the upper the upper an an an lower lower semicircles of of the of the circle the circle x 2 x 2 x 2 y 2 y 2 y (See 25. (See. (See Figure Figure 1.) 1.) 1.) y y y y y y y y y x x x x x x x x x 1 (a) (a) + =25 (a) (b) (b) ƒ=œ (b) (c) (c) =_œ (c) It s It s not It s not easy not easy easy to to solve to solve Equation Equation 2 for 2 for 2 yfor explicitly y y explicitly as as a as function a a function of xof by x by xhan. by han. (A (A computeputer algebra algebra system system has has has no no trouble, no trouble, but but the but the expressions the it it obtains it obtains are are very are very very compli- compli- (A com- The graphs of f an g are the upper an lower semicircles of the circle x cate.) 2 + y cate.) Nonetheless, 2 = 25. (2) (2) is (2) is the is the equation the equation of of a of curve, a a curve, calle calle the the folium the folium of of Descartes, of MAT 1001 Calculus I 43 / 66
44 Implicit Differentiation It s not easy to solve Equation (4) for y explicitly as a function of x by han. Nonetheless, (4) is the equation of a curve, calle the folium of Descartes shown in figure y +Á=6xy three suc implicitl is true fo 0 x FIGURE 2 The folium of Descartes FIGURE 3 MAT 1001 Calculus I 44 / 66
45 implicitly by Equation by Equation by 2, we 2, mean we 2, we mean that mean that the that equation the the equation Implicit Differentiation x 3 x 3 f x x 3 f 3 x f x 3 6xf x 3 6xf x 6xf x an it implicitly efines y as several functions of x. true is true for is true all for values for all values all of values xof in xthe of in xomain the the omain omain of fof. fof. f. y y y y y y y y y x x x x x x x x x IGURE s FIGURE FIGURE 3 Graphs 3 Graphs 3 of Graphs three of three functions of three functions functions efine efine by efine the by folium the by folium the of folium Descartes of Descartes of When we say that f is a function efine implicitly by Equation (4), we Fortunately, mean that Fortunately, we theon t we equation we on t nee on t nee to nee solve to solve to an solve equation an equation an for yfor in for yterms in yterms in of terms xof in xorer of in xorer in to orer fin to fin to fin e erivative the the erivative of y. of Instea yof. Instea y. Instea we can we we can use can use the use metho the the metho metho of implicit of implicit of implicit ifferentiation. ifferentiation. This This Thi onsists consists consists of ifferentiating of of ifferentiating both both sies xboth 3 sies + of [f(x)] sies the of equation the of 3 the = equation 6xf(x) with with respect with respect respect to xto an xto an then x an then solvg the ing resulting the the resulting equation equation for y for. In for y. the y In. examples the In the examples an an exercises an exercises of this of this section of this section section it isit isit i then solv- solv lways isalways true always assume for assume allthat values that the that given the of the xgiven in equation given the equation omain etermines of etermines f. y implicitly y implicitly y as a as ifferentiable a as a ifferentiabl unction function function of xof so xof that so xthat the so that metho the the metho metho of implicit of implicit of implicit ifferentiation ifferentiation can can be applie. can be applie. be applie. XAMPLE EXAMPLE EXAMPLE MAT 1001 Calculus I 45 / 66
46 Implicit Differentiation Fortunately, we on t nee to solve an equation for y in terms of x in orer to fin the erivative of y. Instea we can use the metho of implicit ifferentiation. This consists of ifferentiating both sies of the equation with respect to x an then solving the resulting equation for y. In the examples an exercises of this section it is always assume that the given equation etermines y implicitly as a ifferentiable function of x so that the metho of implicit ifferentiation can be applie. MAT 1001 Calculus I 46 / 66
47 Implicit Differentiation Example 22 (a) If x 2 + y 2 = 25, fin y x. (b) Fin an equation of the tangent to the circle x 2 + y 2 = 25 at the point (3, 4). Solution 1. (a) Differentiate both sies of the equation x 2 + y 2 = 25: x (x2 + y 2 ) = x (25) x (x2 ) + x (y2 ) = 0 Remembering that y is a function of x an using the Chain Rule, we have x (y2 ) = y (y2 ) y y = 2y x x. MAT 1001 Calculus I 47 / 66
48 Implicit Differentiation Solution 1 (cont.) Thus 2x + 2y y x = 0. Now we solve this equation for y/x: y x = x y. (b) At the point (3, 4) we have x = 3 an y = 4, so y x = 3 4 An equation of the tangent to the circle at (3, 4) is therefore y 4 = 3 (x 3) or 3x + 4y = MAT 1001 Calculus I 48 / 66
49 Implicit Differentiation Solution 2. Solving the equation x 2 + y 2 = 25, we get y = ± 25 x 2. The point (3, 4) lies on the upper semicircle y = 25 x 2 an so we consier the function f(x) = y = 25 x 2. Differentiating f using the Chain Rule, we have f (x) = 1 2 (25 x2 ) 1/2 x (25 x2 ) = 1 2 (25 x2 ) 1/2 x ( 2x) = 25 x 2 So f 3 (3) = = tangent is 3x + 4y = 25. an, as in Solution 1, an equation of the MAT 1001 Calculus I 49 / 66
50 Implicit Differentiation Note 1 Previous example illustrates that even when it is possible to solve an equation explicitly for y in terms of x, it may be easier to use implicit ifferentiation. Note 2 The expression y/x = x/y gives the erivative in terms of both x an y. It is correct no matter which function y is etermine by the given equation. For instance, for y = f(x) = 25 x 2 we have y x = x y = x. 25 x 2 whereas for y = g(x) = 25 x 2 we have y x = x y = x 25 x 2 = x 25 x 2. MAT 1001 Calculus I 50 / 66
51 Implicit Differentiation Example 23 (a) Fin y if x 3 + y 3 = 6xy. (b) Fin the tangent to the folium of Descartes x 3 + y 3 = 6xy at the point (3, 3). Solution. (a) Differentiating both sies of x 3 + y 3 = 6xy with respect to x, regaring y as a function of x, an using the Chain Rule on the y 3 term an the Prouct Rule on the 6xy term, we get 3x 2 + 3y 2 y = 6y + 6xy or x 2 + y 2 y = 2y + 2xy. MAT 1001 Calculus I 51 / 66
52 Implicit Differentiation Solution (cont.) We now solve for y : (b) When x = y = 3, y 2 y 2xy = 2y x 2 (y 2 2x)y = 2y x 2 y = 2y x2 y 2 2x y = = 1 So an equation of the tangent to the folium at (3, 3) is y 3 = 1(x 3) or x + y = 6. MAT 1001 Calculus I 52 / 66
53 Implicit Differentiation Derivatives of Inverse Trigonometric Functions Derivatives of Inverse Trigonometric Functions We can use implicit ifferentiation to fin the erivatives of the inverse trigonometric functions, assuming that these functions are ifferentiable. Recall the efinition of the arcsine function: y = sin 1 x sin y = x an π 2 y π 2. Differentiating sin y = x implicitly with respect to x, we obtain cos y y x = 1 or y x = 1 cos y. MAT 1001 Calculus I 53 / 66
54 Implicit Differentiation Derivatives of Inverse Trigonometric Functions (arcsin x) We know y x = 1. Now cos y 0, since π/2 y π/2, so cos y cos y = 1 sin 2 y = 1 x 2. Therefore y x = 1 cos y = 1 1 x 2 (arctan x) x (sin 1 x) = 1 1 x 2. The formula for the erivative of the arctangent function is erive in a similar way. x (tan( 1) ) = x 2. MAT 1001 Calculus I 54 / 66
55 Implicit Differentiation Derivatives of Inverse Trigonometric Functions Example 24 Differentiate (a) y = Solution. (a) 1 sin 1 x an (b) f(x) = x arctan x. y x = x (sin 1 x) 1 = (sin 1 x) 2 x (sin 1 x) 1 = (sin 1 x) 2 1 x 2 (b) f 1 (x) = x 1 + ( x) 2 x = 2(1 + x) + arctan x ( 1 2 x 1/2 ) + arctan x The inverse trigonometric functions that occur most frequently are the ones that we have just iscusse. MAT 1001 Calculus I 55 / 66
56 Derivatives of Logarithmic Functions Derivatives of Logarithmic Functions In this section we use implicit ifferentiation to fin the erivatives of the logarithmic functions y = log a x an, in particular, the natural logarithmic function y = ln x. x (log a x) = 1 (5) x ln a If we put a = e, we have x (ln x) = 1 x. (6) In general, if we combine Formula (6) with the Chain Rule as in previous example, we get x (ln u) = 1 u u x or x (ln g(x)) = g (x) g(x) (7) MAT 1001 Calculus I 56 / 66
57 Derivatives of Logarithmic Functions Example 25 Differentiate y = ln(x 3 + 1). Solution. To use the Chain Rule we let u = x Then y = ln u, so y x = y u u x = 1 u u x = 1 x (3x2 ) = 3x2 x MAT 1001 Calculus I 57 / 66
58 Derivatives of Logarithmic Functions Example 26 Differentiate f(x) = log 10 (2 + sin x). Solution. Using Formula (5) with a = 10, we have f (x) = x log 10(2 + sin x) = cos x = (2 + sin x) ln (2 + sin x) (2 + sin x) ln 10 x MAT 1001 Calculus I 58 / 66
59 Derivatives of Logarithmic Functions Example 27 Fin f (x) if f(x) = ln x. Solution. Since it follows that Thus, f (x) = 1 x f(x) = { ln x, x > 0 ln( x), x < 0 1 x, x > 0 f (x) = 1 x ( 1) = 1 x, x < 0 for all x 0. MAT 1001 Calculus I 59 / 66
60 Derivatives of Logarithmic Functions Logarithmic Differentiation Logarithmic Differentiation Example 28 Differentiate y = x3/4 x (3x + 2) 2. Solution. We take logarithms of both sies of the equation an use the Laws of Logarithms to simplify: ln y = 3 4 ln x ln(x2 + 1) 5 ln(3x + 2) Differentiating implicitly with respect to x gives y x y = x x x x + 2 MAT 1001 Calculus I 60 / 66
61 Derivatives of Logarithmic Functions Logarithmic Differentiation Solution. Solving for y/x, we get y x = y y x y = 3 4x + x x x + 2 ( 3 4x + = x3/4 x (3x + 2) 5 x x ) 3x + 2 ( 3 4x + x x x + 2 ) MAT 1001 Calculus I 61 / 66
62 Derivatives of Logarithmic Functions Logarithmic Differentiation Note You shoul istinguish carefully between the Power Rule [(x n ) = nx n 1 ], where the base is variable an the exponent is constant, an the rule for ifferentiating exponential functions [(a x ) = a x ln a], where the base is constant an the exponent is variable. In general there are four cases for exponents an bases: 1 x (ab ) = 0 (a an b are constants.) 2 x [f(x)b ] = b[f(x)] b 1 f (x) 3 x [ag(x) ] = a g(x) (ln a)g (x) 4 To fin x [f(x)]g(x), logarithmic ifferentiation can be use, as in the next example. MAT 1001 Calculus I 62 / 66
63 Derivatives of Logarithmic Functions Logarithmic Differentiation Example 29 Differentiate y = x x. Solution 1. Using logarithmic ifferentiation, we have ln y = ln x x = x ln x y y = x 1 x + (ln x) 1 2 x ( 1 y = y x + ln x ) 2 x ( ) = x x 2 + ln x 2. x MAT 1001 Calculus I 63 / 66
64 Derivatives of Logarithmic Functions Logarithmic Differentiation Solution 2. Another metho is to write x x = ( e ln x) x : ( x ) x = (e ) x ln x x x = e x ln x x ( x ln x) ( ) = x x 2 + ln x 2 x MAT 1001 Calculus I 64 / 66
65 Derivatives of Logarithmic Functions The Number e as a Limit The Number e as a Limit We have shown that if f(x) = ln x, then f (x) = 1/x. Thus, f (1) = 1. We now use this fact to express the number e as a limit. From the efinition of a erivative as a limit, we have f f(1 + h) f(1) f(1 + x) f(1) (1) = lim = lim h 0 h x 0 x = lim x 0 ln(1 + x) ln 1 x 1 = lim ln(1 + x) x 0 x = lim ln(1 + x) 1/x x 0 [ = ln lim (1 + x)1/x] (since ln is continuous) x 0 MAT 1001 Calculus I 65 / 66
66 Derivatives of Logarithmic Functions The Number e as a Limit Because f (1) = 1, we have Therefore ln [ lim x 0 (1 + x)1/x] = 1. lim (1 + x 0 x)1/x = e. (8) If we put n = 1/x in Formula (8), then n as x 0 + an so an alternative expression for e is ( e = lim n. n n) MAT 1001 Calculus I 66 / 66
February 21 Math 1190 sec. 63 Spring 2017
February 21 Math 1190 sec. 63 Spring 2017 Chapter 2: Derivatives Let s recall the efinitions an erivative rules we have so far: Let s assume that y = f (x) is a function with c in it s omain. The erivative
More informationMath 115 Section 018 Course Note
Course Note 1 General Functions Definition 1.1. A function is a rule that takes certain numbers as inputs an assigns to each a efinite output number. The set of all input numbers is calle the omain of
More informationDIFFERENTIATION RULES
3 DIFFERENTIATION RULES DIFFERENTIATION RULES The functions that we have met so far can be described by expressing one variable explicitly in terms of another variable. y For example,, or y = x sin x,
More informationMath Chapter 2 Essentials of Calculus by James Stewart Prepared by Jason Gaddis
Math 231 - Chapter 2 Essentials of Calculus by James Stewart Prepare by Jason Gais Chapter 2 - Derivatives 21 - Derivatives an Rates of Change Definition A tangent to a curve is a line that intersects
More information1 Lecture 18: The chain rule
1 Lecture 18: The chain rule 1.1 Outline Comparing the graphs of sin(x) an sin(2x). The chain rule. The erivative of a x. Some examples. 1.2 Comparing the graphs of sin(x) an sin(2x) We graph f(x) = sin(x)
More informationImplicit Differentiation
Implicit Differentiation Thus far, the functions we have been concerne with have been efine explicitly. A function is efine explicitly if the output is given irectly in terms of the input. For instance,
More informationSection The Chain Rule and Implicit Differentiation with Application on Derivative of Logarithm Functions
Section 3.4-3.6 The Chain Rule an Implicit Differentiation with Application on Derivative of Logarithm Functions Ruipeng Shen September 3r, 5th Ruipeng Shen MATH 1ZA3 September 3r, 5th 1 / 3 The Chain
More information1 Lecture 20: Implicit differentiation
Lecture 20: Implicit ifferentiation. Outline The technique of implicit ifferentiation Tangent lines to a circle Derivatives of inverse functions by implicit ifferentiation Examples.2 Implicit ifferentiation
More informationf(x) f(a) Limit definition of the at a point in slope notation.
Lesson 9: Orinary Derivatives Review Hanout Reference: Brigg s Calculus: Early Transcenentals, Secon Eition Topics: Chapter 3: Derivatives, p. 126-235 Definition. Limit Definition of Derivatives at a point
More information1 Definition of the derivative
Math 20A - Calculus by Jon Rogawski Chapter 3 - Differentiation Prepare by Jason Gais Definition of the erivative Remark.. Recall our iscussion of tangent lines from way back. We now rephrase this in terms
More informationd dx [xn ] = nx n 1. (1) dy dx = 4x4 1 = 4x 3. Theorem 1.3 (Derivative of a constant function). If f(x) = k and k is a constant, then f (x) = 0.
Calculus refresher Disclaimer: I claim no original content on this ocument, which is mostly a summary-rewrite of what any stanar college calculus book offers. (Here I ve use Calculus by Dennis Zill.) I
More informationThe Chain Rule. Composition Review. Intuition. = 2(1.5) = 3 times faster than (X)avier.
The Chain Rule In the previous section we ha to use a trig ientity to etermine the erivative of. h(x) = sin(2x). We can view h(x) as the composition of two functions. Let g(x) = 2x an f (x) = sin x. Then
More informationLecture 4 : General Logarithms and Exponentials. a x = e x ln a, a > 0.
For a > 0 an x any real number, we efine Lecture 4 : General Logarithms an Exponentials. a x = e x ln a, a > 0. The function a x is calle the exponential function with base a. Note that ln(a x ) = x ln
More informationChapter 3 Differentiation Rules
Chapter 3 Differentiation Rules Derivative constant function if c is any real number, then Example: The Power Rule: If n is a positive integer, then Example: Extended Power Rule: If r is any real number,
More informationMath Implicit Differentiation. We have discovered (and proved) formulas for finding derivatives of functions like
Math 400 3.5 Implicit Differentiation Name We have iscovere (an prove) formulas for fining erivatives of functions like f x x 3x 4x. 3 This amounts to fining y for 3 y x 3x 4x. Notice that in this case,
More informationDifferentiability, Computing Derivatives, Trig Review. Goals:
Secants vs. Derivatives - Unit #3 : Goals: Differentiability, Computing Derivatives, Trig Review Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an
More informationImplicit Differentiation and Inverse Trigonometric Functions
Implicit Differentiation an Inverse Trigonometric Functions MATH 161 Calculus I J. Robert Buchanan Department of Mathematics Summer 2018 Explicit vs. Implicit Functions 0.5 1 y 0.0 y 2 0.5 3 4 1.0 0.5
More informationChapter 3 Notes, Applied Calculus, Tan
Contents 3.1 Basic Rules of Differentiation.............................. 2 3.2 The Prouct an Quotient Rules............................ 6 3.3 The Chain Rule...................................... 9 3.4
More informationImplicit Differentiation
Implicit Differentiation Implicit Differentiation Using the Chain Rule In the previous section we focuse on the erivatives of composites an saw that THEOREM 20 (Chain Rule) Suppose that u = g(x) is ifferentiable
More information23 Implicit differentiation
23 Implicit ifferentiation 23.1 Statement The equation y = x 2 + 3x + 1 expresses a relationship between the quantities x an y. If a value of x is given, then a corresponing value of y is etermine. For
More informationDifferentiation ( , 9.5)
Chapter 2 Differentiation (8.1 8.3, 9.5) 2.1 Rate of Change (8.2.1 5) Recall that the equation of a straight line can be written as y = mx + c, where m is the slope or graient of the line, an c is the
More informationSome functions and their derivatives
Chapter Some functions an their erivatives. Derivative of x n for integer n Recall, from eqn (.6), for y = f (x), Also recall that, for integer n, Hence, if y = x n then y x = lim δx 0 (a + b) n = a n
More informationReview of Differentiation and Integration for Ordinary Differential Equations
Schreyer Fall 208 Review of Differentiation an Integration for Orinary Differential Equations In this course you will be expecte to be able to ifferentiate an integrate quickly an accurately. Many stuents
More informationTOTAL NAME DATE PERIOD AP CALCULUS AB UNIT 4 ADVANCED DIFFERENTIATION TECHNIQUES DATE TOPIC ASSIGNMENT /6 10/8 10/9 10/10 X X X X 10/11 10/12
NAME DATE PERIOD AP CALCULUS AB UNIT ADVANCED DIFFERENTIATION TECHNIQUES DATE TOPIC ASSIGNMENT 0 0 0/6 0/8 0/9 0/0 X X X X 0/ 0/ 0/5 0/6 QUIZ X X X 0/7 0/8 0/9 0/ 0/ 0/ 0/5 UNIT EXAM X X X TOTAL AP Calculus
More informationCalculus I Practice Test Problems for Chapter 3 Page 1 of 9
Calculus I Practice Test Problems for Chapter 3 Page of 9 This is a set of practice test problems for Chapter 3. This is in no wa an inclusive set of problems there can be other tpes of problems on the
More information016A Homework 10 Solution
016A Homework 10 Solution Jae-young Park November 2, 2008 4.1 #14 Write each expression in the form of 2 kx or 3 kx, for a suitable constant k; (3 x 3 x/5 ) 5, (16 1/4 16 3/4 ) 3x Solution (3 x 3 x/5 )
More informationDifferentiability, Computing Derivatives, Trig Review
Unit #3 : Differentiability, Computing Derivatives, Trig Review Goals: Determine when a function is ifferentiable at a point Relate the erivative graph to the the graph of an original function Compute
More informationSYDE 112, LECTURE 1: Review & Antidifferentiation
SYDE 112, LECTURE 1: Review & Antiifferentiation 1 Course Information For a etaile breakown of the course content an available resources, see the Course Outline. Other relevant information for this section
More informationFinal Exam Study Guide and Practice Problems Solutions
Final Exam Stuy Guie an Practice Problems Solutions Note: These problems are just some of the types of problems that might appear on the exam. However, to fully prepare for the exam, in aition to making
More informationOutline. MS121: IT Mathematics. Differentiation Rules for Differentiation: Part 1. Outline. Dublin City University 4 The Quotient Rule
MS2: IT Mathematics Differentiation Rules for Differentiation: Part John Carroll School of Mathematical Sciences Dublin City University Pattern Observe You may have notice the following pattern when we
More informationBy writing (1) as y (x 5 1). (x 5 1), we can find the derivative using the Product Rule: y (x 5 1) 2. we know this from (2)
3.5 Chain Rule 149 3.5 Chain Rule Introuction As iscusse in Section 3.2, the Power Rule is vali for all real number exponents n. In this section we see that a similar rule hols for the erivative of a power
More informationDefine each term or concept.
Chapter Differentiation Course Number Section.1 The Derivative an the Tangent Line Problem Objective: In this lesson you learne how to fin the erivative of a function using the limit efinition an unerstan
More informationThe derivative of a function f(x) is another function, defined in terms of a limiting expression: f(x + δx) f(x)
Y. D. Chong (2016) MH2801: Complex Methos for the Sciences 1. Derivatives The erivative of a function f(x) is another function, efine in terms of a limiting expression: f (x) f (x) lim x δx 0 f(x + δx)
More informationAP Calculus AB One Last Mega Review Packet of Stuff. Take the derivative of the following. 1.) 3.) 5.) 7.) Determine the limit of the following.
AP Calculus AB One Last Mega Review Packet of Stuff Name: Date: Block: Take the erivative of the following. 1.) x (sin (5x)).) x (etan(x) ) 3.) x (sin 1 ( x3 )) 4.) x (x3 5x) 4 5.) x ( ex sin(x) ) 6.)
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 information1 Limits Finding limits graphically. 1.3 Finding limits analytically. Examples 1. f(x) = x3 1. f(x) = f(x) =
Theorem 13 (i) If p(x) is a polynomial, then p(x) = p(c) 1 Limits 11 12 Fining its graphically Examples 1 f(x) = x3 1, x 1 x 1 The behavior of f(x) as x approximates 1 x 1 f(x) = 3 x 2 f(x) = x+1 1 f(x)
More informationChapter 2. Exponential and Log functions. Contents
Chapter. Exponential an Log functions This material is in Chapter 6 of Anton Calculus. The basic iea here is mainly to a to the list of functions we know about (for calculus) an the ones we will stu all
More informationMath 180, Exam 2, Fall 2012 Problem 1 Solution. (a) The derivative is computed using the Chain Rule twice. 1 2 x x
. Fin erivatives of the following functions: (a) f() = tan ( 2 + ) ( ) 2 (b) f() = ln 2 + (c) f() = sin() Solution: Math 80, Eam 2, Fall 202 Problem Solution (a) The erivative is compute using the Chain
More informationcosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) = 2 2t 1 + t 2 cos x = 1 t2 We will revisit the double angle identities:
6.4 Integration using tanx/) We will revisit the ouble angle ientities: sin x = sinx/) cosx/) = tanx/) sec x/) = tanx/) + tan x/) cos x = cos x/) sin x/) tan x = = tan x/) sec x/) tanx/) tan x/). = tan
More informationSolutions to Math 41 Second Exam November 4, 2010
Solutions to Math 41 Secon Exam November 4, 2010 1. (13 points) Differentiate, using the metho of your choice. (a) p(t) = ln(sec t + tan t) + log 2 (2 + t) (4 points) Using the rule for the erivative of
More informationExam 3 Review. Lesson 19: Concavity, Inflection Points, and the Second Derivative Test. Lesson 20: Absolute Extrema on an Interval
Exam 3 Review Lessons 17-18: Relative Extrema, Critical Numbers, an First Derivative Test (from exam 2 review neee for curve sketching) Critical Numbers: where the erivative of a function is zero or unefine.
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 informationChapter 2 Derivatives
Chapter Derivatives Section. An Intuitive Introuction to Derivatives Consier a function: Slope function: Derivative, f ' For each, the slope of f is the height of f ' Where f has a horizontal tangent line,
More informationARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT
ARAB ACADEMY FOR SCIENCE TECHNOLOGY AND MARITIME TRANSPORT Course: Math For Engineering Winter 8 Lecture Notes By Dr. Mostafa Elogail Page Lecture [ Functions / Graphs of Rational Functions] Functions
More informationSECTION 3.2 THE PRODUCT AND QUOTIENT RULES 1 8 3
SECTION 3.2 THE PRODUCT AND QUOTIENT RULES 8 3 L P f Q L segments L an L 2 to be tangent to the parabola at the transition points P an Q. (See the figure.) To simplify the equations you ecie to place the
More informationJUST THE MATHS UNIT NUMBER DIFFERENTIATION 2 (Rates of change) A.J.Hobson
JUST THE MATHS UNIT NUMBER 10.2 DIFFERENTIATION 2 (Rates of change) by A.J.Hobson 10.2.1 Introuction 10.2.2 Average rates of change 10.2.3 Instantaneous rates of change 10.2.4 Derivatives 10.2.5 Exercises
More informationMath 190 Chapter 3 Lecture Notes. Professor Miguel Ornelas
Math 190 Chapter 3 Lecture Notes Professor Miguel Ornelas 1 M. Ornelas Math 190 Lecture Notes Section 3.1 Section 3.1 Derivatives of Polynomials an Exponential Functions Derivative of a Constant Function
More informationdx dx [x2 + y 2 ] = y d [tan x] + tan x = 2x + 2y = y sec 2 x + tan x dy dy = tan x dy dy = [tan x 2y] dy dx = 2x y sec2 x [1 + sin y] = sin(xy)
Math 7 Activit: Implicit & Logarithmic Differentiation (Solutions) Implicit Differentiation. For each of the following equations, etermine x. a. tan x = x 2 + 2 tan x] = x x x2 + 2 ] = tan x] + tan x =
More informationMath 210 Midterm #1 Review
Math 20 Miterm # Review This ocument is intene to be a rough outline of what you are expecte to have learne an retaine from this course to be prepare for the first miterm. : Functions Definition: A function
More informationCalculus I Announcements
Slie 1 Calculus I Announcements Office Hours: Amos Eaton 309, Monays 12:50-2:50 Exam 2 is Thursay, October 22n. The stuy guie is now on the course web page. Start stuying now, an make a plan to succee.
More informationMath 1B, lecture 8: Integration by parts
Math B, lecture 8: Integration by parts Nathan Pflueger 23 September 2 Introuction Integration by parts, similarly to integration by substitution, reverses a well-known technique of ifferentiation an explores
More informationLinear First-Order Equations
5 Linear First-Orer Equations Linear first-orer ifferential equations make up another important class of ifferential equations that commonly arise in applications an are relatively easy to solve (in theory)
More informationMath 251 Notes. Part I.
Math 251 Notes. Part I. F. Patricia Meina May 6, 2013 Growth Moel.Consumer price inex. [Problem 20, page 172] The U.S. consumer price inex (CPI) measures the cost of living base on a value of 100 in the
More informationUNIT 3: DERIVATIVES STUDY GUIDE
Calculus I UNIT 3: Derivatives REVIEW Name: Date: UNIT 3: DERIVATIVES STUDY GUIDE Section 1: Section 2: Limit Definition (Derivative as the Slope of the Tangent Line) Calculating Rates of Change (Average
More informationL Hôpital s Rule was discovered by Bernoulli but written for the first time in a text by L Hôpital.
7.5. Ineterminate Forms an L Hôpital s Rule L Hôpital s Rule was iscovere by Bernoulli but written for the first time in a text by L Hôpital. Ineterminate Forms 0/0 an / f(x) If f(x 0 ) = g(x 0 ) = 0,
More informationFUNCTIONS AND MODELS
1 FUNCTIONS AND MODELS FUNCTIONS AND MODELS 1.6 Inverse Functions and Logarithms In this section, we will learn about: Inverse functions and logarithms. INVERSE FUNCTIONS The table gives data from an experiment
More informationYou should also review L Hôpital s Rule, section 3.6; follow the homework link above for exercises.
BEFORE You Begin Calculus II If it has been awhile since you ha Calculus, I strongly suggest that you refresh both your ifferentiation an integration skills. I woul also like to remin you that in Calculus,
More informationInverse Trig Functions
Inverse Trig Functions -8-006 If you restrict fx) = sinx to the interval π x π, the function increases: y = sin x - / / This implies that the function is one-to-one, an hence it has an inverse. The inverse
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 informationFall 2016: Calculus I Final
Answer the questions in the spaces provie on the question sheets. If you run out of room for an answer, continue on the back of the page. NO calculators or other electronic evices, books or notes are allowe
More informationChapter 1. Functions, Graphs, and Limits
Review for Final Exam Lecturer: Sangwook Kim Office : Science & Tech I, 226D math.gmu.eu/ skim22 Chapter 1. Functions, Graphs, an Limits A function is a rule that assigns to each objects in a set A exactly
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 informationYour signature: (1) (Pre-calculus Review Set Problems 80 and 124.)
(1) (Pre-calculus Review Set Problems 80 an 14.) (a) Determine if each of the following statements is True or False. If it is true, explain why. If it is false, give a counterexample. (i) If a an b are
More information3 Algebraic Methods. we can differentiate both sides implicitly to obtain a differential equation involving x and y:
3 Algebraic Methods b The first appearance of the equation E Mc 2 in Einstein s handwritten notes. So far, the only general class of differential equations that we know how to solve are directly integrable
More informationMath 147 Exam II Practice Problems
Math 147 Exam II Practice Problems This review should not be used as your sole source for preparation for the exam. You should also re-work all examples given in lecture, all homework problems, all lab
More informationQF101: Quantitative Finance September 5, Week 3: Derivatives. Facilitator: Christopher Ting AY 2017/2018. f ( x + ) f(x) f(x) = lim
QF101: Quantitative Finance September 5, 2017 Week 3: Derivatives Facilitator: Christopher Ting AY 2017/2018 I recoil with ismay an horror at this lamentable plague of functions which o not have erivatives.
More informationMath 1271 Solutions for Fall 2005 Final Exam
Math 7 Solutions for Fall 5 Final Eam ) Since the equation + y = e y cannot be rearrange algebraically in orer to write y as an eplicit function of, we must instea ifferentiate this relation implicitly
More informationLecture 5: Inverse Trigonometric Functions
Lecture 5: Inverse Trigonometric Functions 5 The inverse sine function The function f(x = sin(x is not one-to-one on (,, but is on [ π, π Moreover, f still has range [, when restricte to this interval
More informationIMPLICIT DIFFERENTIATION
IMPLICIT DIFFERENTIATION CALCULUS 3 INU0115/515 (MATHS 2) Dr Arian Jannetta MIMA CMath FRAS Implicit Differentiation 1/ 11 Arian Jannetta Explicit an implicit functions Explicit functions An explicit function
More informationDERIVATIVES: LAWS OF DIFFERENTIATION MR. VELAZQUEZ AP CALCULUS
DERIVATIVES: LAWS OF DIFFERENTIATION MR. VELAZQUEZ AP CALCULUS THE DERIVATIVE AS A FUNCTION f x = lim h 0 f x + h f(x) h Last class we examine the limit of the ifference quotient at a specific x as h 0,
More informationFurther Differentiation and Applications
Avance Higher Notes (Unit ) Prerequisites: Inverse function property; prouct, quotient an chain rules; inflexion points. Maths Applications: Concavity; ifferentiability. Real-Worl Applications: Particle
More informationInverse Trig Functions
Inverse Trig Functions -7-08 If you restrict fx) = sinx to the interval π x π, the function increases: y = sin x - / / This implies that the function is one-to-one, an hence it has an inverse. The inverse
More informationMA123, Supplement: Exponential and logarithmic functions (pp , Gootman)
MA23, Supplement: Exponential an logarithmic functions pp. 35-39, Gootman) Chapter Goals: Review properties of exponential an logarithmic functions. Learn how to ifferentiate exponential an logarithmic
More informationd dx But have you ever seen a derivation of these results? We ll prove the first result below. cos h 1
Lecture 5 Some ifferentiation rules Trigonometric functions (Relevant section from Stewart, Seventh Eition: Section 3.3) You all know that sin = cos cos = sin. () But have you ever seen a erivation of
More information1 Lecture 13: The derivative as a function.
1 Lecture 13: Te erivative as a function. 1.1 Outline Definition of te erivative as a function. efinitions of ifferentiability. Power rule, erivative te exponential function Derivative of a sum an a multiple
More information11.7. Implicit Differentiation. Introduction. Prerequisites. Learning Outcomes
Implicit Differentiation 11.7 Introuction This Section introuces implicit ifferentiation which is use to ifferentiate functions expresse in implicit form (where the variables are foun together). Examples
More informationFind the indicated derivative. 1) Find y(4) if y = 3 sin x. A) y(4) = 3 cos x B) y(4) = 3 sin x C) y(4) = - 3 cos x D) y(4) = - 3 sin x
Assignment 5 Name Find the indicated derivative. ) Find y(4) if y = sin x. ) A) y(4) = cos x B) y(4) = sin x y(4) = - cos x y(4) = - sin x ) y = (csc x + cot x)(csc x - cot x) ) A) y = 0 B) y = y = - csc
More informationMake graph of g by adding c to the y-values. on the graph of f by c. multiplying the y-values. even-degree polynomial. graph goes up on both sides
Reference 1: Transformations of Graphs an En Behavior of Polynomial Graphs Transformations of graphs aitive constant constant on the outsie g(x) = + c Make graph of g by aing c to the y-values on the graph
More informationTest one Review Cal 2
Name: Class: Date: ID: A Test one Review Cal 2 Short Answer. Write the following expression as a logarithm of a single quantity. lnx 2ln x 2 ˆ 6 2. Write the following expression as a logarithm of a single
More informationMath 106 Exam 2 Topics. du dx
The Chain Rule Math 106 Exam 2 Topics Composition (g f)(x 0 ) = g(f(x 0 )) ; (note: we on t know what g(x 0 ) is.) (g f) ought to have something to o with g (x) an f (x) in particular, (g f) (x 0 ) shoul
More informationChapter 4 Notes, Calculus I with Precalculus 3e Larson/Edwards
4.1 The Derivative Recall: For the slope of a line we need two points (x 1,y 1 ) and (x 2,y 2 ). Then the slope is given by the formula: m = y x = y 2 y 1 x 2 x 1 On a curve we can find the slope of a
More information7.1. Calculus of inverse functions. Text Section 7.1 Exercise:
Contents 7. Inverse functions 1 7.1. Calculus of inverse functions 2 7.2. Derivatives of exponential function 4 7.3. Logarithmic function 6 7.4. Derivatives of logarithmic functions 7 7.5. Exponential
More informationTable of Contents Derivatives of Logarithms
Derivatives of Logarithms- Table of Contents Derivatives of Logarithms Arithmetic Properties of Logarithms Derivatives of Logarithms Example Example 2 Example 3 Example 4 Logarithmic Differentiation Example
More informationUsing the definition of the derivative of a function is quite tedious. f (x + h) f (x)
Derivative Rules Using te efinition of te erivative of a function is quite teious. Let s prove some sortcuts tat we can use. Recall tat te efinition of erivative is: Given any number x for wic te limit
More informationAnnouncements. Topics: Homework: - sections 4.5 and * Read these sections and study solved examples in your textbook!
Announcements Topics: - sections 4.5 and 5.1-5.5 * Read these sections and study solved examples in your textbook! Homework: - review lecture notes thoroughly - work on practice problems from the textbook
More informationMATH 120 Theorem List
December 11, 2016 Disclaimer: Many of the theorems covere in class were not name, so most of the names on this sheet are not efinitive (they are escriptive names rather than given names). Lecture Theorems
More informationMAT01A1: Differentiation of Polynomials & Exponential Functions + the Product & Quotient Rules
MAT01A1: Differentiation of Polynomials & Exponential Functions + te Prouct & Quotient Rules Dr Craig 22 Marc 2017 Semester Test 1 Scripts will be available for collection from Tursay morning. For marking
More informationTHEOREM: THE CONSTANT RULE
MATH /MYERS/ALL FORMULAS ON THIS REVIEW MUST BE MEMORIZED! DERIVATIVE REVIEW THEOREM: THE CONSTANT RULE The erivative of a constant function is zero. That is, if c is a real number, then c 0 Eample 1:
More informationTangent Lines Sec. 2.1, 2.7, & 2.8 (continued)
Tangent Lines Sec. 2.1, 2.7, & 2.8 (continued) Prove this Result How Can a Derivative Not Exist? Remember that the derivative at a point (or slope of a tangent line) is a LIMIT, so it doesn t exist whenever
More informationHyperbolic Functions. Notice: this material must not be used as a substitute for attending. the lectures
Hyperbolic Functions Notice: this material must not be use as a substitute for attening the lectures 0. Hyperbolic functions sinh an cosh The hyperbolic functions sinh (pronounce shine ) an cosh are efine
More information3.4 The Chain Rule. F (x) = f (g(x))g (x) Alternate way of thinking about it: If y = f(u) and u = g(x) where both are differentiable functions, then
3.4 The Chain Rule To find the derivative of a function that is the composition of two functions for which we already know the derivatives, we can use the Chain Rule. The Chain Rule: Suppose F (x) = f(g(x)).
More informationLecture 6: Calculus. In Song Kim. September 7, 2011
Lecture 6: Calculus In Song Kim September 7, 20 Introuction to Differential Calculus In our previous lecture we came up with several ways to analyze functions. We saw previously that the slope of a linear
More informationSTUDENT S COMPANIONS IN BASIC MATH: THE FOURTH. Trigonometric Functions
STUDENT S COMPANIONS IN BASIC MATH: THE FOURTH Trigonometric Functions Let me quote a few sentences at the beginning of the preface to a book by Davi Kammler entitle A First Course in Fourier Analysis
More informationImplicit Differentiation. Lecture 16.
Implicit Differentiation. Lecture 16. We are use to working only with functions that are efine explicitly. That is, ones like f(x) = 5x 3 + 7x x 2 + 1 or s(t) = e t5 3, in which the function is escribe
More informationFlash Card Construction Instructions
Flash Car Construction Instructions *** THESE CARDS ARE FOR CALCULUS HONORS, AP CALCULUS AB AND AP CALCULUS BC. AP CALCULUS BC WILL HAVE ADDITIONAL CARDS FOR THE COURSE (IN A SEPARATE FILE). The left column
More informationMath 1 Lecture 20. Dartmouth College. Wednesday
Math 1 Lecture 20 Dartmouth College Wenesay 10-26-16 Contents Reminers/Announcements Last Time Derivatives of Trigonometric Functions Reminers/Announcements WebWork ue Friay x-hour problem session rop
More informationA. Incorrect! The letter t does not appear in the expression of the given integral
AP Physics C - Problem Drill 1: The Funamental Theorem of Calculus Question No. 1 of 1 Instruction: (1) Rea the problem statement an answer choices carefully () Work the problems on paper as neee (3) Question
More informationMATH2231-Differentiation (2)
-Differentiation () The Beginnings of Calculus The prime occasion from which arose my iscovery of the metho of the Characteristic Triangle, an other things of the same sort, happene at a time when I ha
More informationcosh x sinh x So writing t = tan(x/2) we have 6.4 Integration using tan(x/2) 2t 1 + t 2 cos x = 1 t2 sin x =
6.4 Integration using tan/ We will revisit the ouble angle ientities: sin = sin/ cos/ = tan/ sec / = tan/ + tan / cos = cos / sin / tan = = tan / sec / tan/ tan /. = tan / + tan / So writing t = tan/ we
More informationModule FP2. Further Pure 2. Cambridge University Press Further Pure 2 and 3 Hugh Neill and Douglas Quadling Excerpt More information
5548993 - Further Pure an 3 Moule FP Further Pure 5548993 - Further Pure an 3 Differentiating inverse trigonometric functions Throughout the course you have graually been increasing the number of functions
More information