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
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
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
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
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
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
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
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
The Prouct an Quotient Rules The Secon Derivative Example 2 x (x8 + 12x 5 4x 4 + 10x 3 6x + 5) = x (x8 ) + 12 x (x5 ) 4 x (x4 ) + 10 x (x3 ) 6 x (x) + x (5) = 8x 7 + 12(5x 4 ) 4(4x 3 ) + 10(3x 2 ) 6(1) + 0 = 8x 7 + 60x 4 16x 3 + 30x 2 6 MAT 1001 Calculus I 9 / 66
The Prouct an Quotient Rules The Secon Derivative Example 3 Fin the points on the curve y = x 4 6x 2 + 4 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
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
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
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
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). 1 2. x 1/2 = x. g (x) + g(x) 2 x So f (4) = 4. g (4) + g(4) 2 4 = 2. 3 + 2 2. 2 = 6.5 MAT 1001 Calculus I 14 / 66
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 3 + 12x + 6) (3x 4 + 3x 3 6x 2 ) (x 3 + 6) 2 = x4 2x 3 + 6x 2 + 12x + 6 (x 3 + 6) 2. MAT 1001 Calculus I 15 / 66
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
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
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
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
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
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
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
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 _2 3.5 0 FIGURE 4 This means [See Figure 1, e 2.] MAT 1001 Calculus I 23 / 66
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
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
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
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
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
The Chain Rule The Chain Rule Suppose you are aske to ifferentiate the function F (x) = x 2 + 1 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
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
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
The Chain Rule Example 14 Fin F (x) if F (x) = x 2 + 1. 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 2 + 1. 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 2 + 1 2x = x x 2 + 1. MAT 1001 Calculus I 32 / 66
The Chain Rule Solution 2. (using Equation (2)): If we let u = x 2 + 1 an y = u, then F (x) = y u u x = 1 2 u 2x 1 = 2 x 2 + 1 2x = x x 2 + 1. MAT 1001 Calculus I 33 / 66
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
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
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
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
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
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
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
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
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 3 + 1 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
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 2 25 25. (See 25. (See. (See Figure Figure 1.) 1.) 1.) y y y y y y y y y 0 0 0 x x x 0 0 0 x x x 0 0 0 x x x 1 (a) (a) + =25 (a) (b) (b) ƒ=œ (b) 25-25- (c) (c) =_œ (c) 25-25- 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
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
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 0 0 0 x x x 0 0 0 x x x 0 0 0 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 1 1 1 MAT 1001 Calculus I 45 / 66
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
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
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 = 25. 4 MAT 1001 Calculus I 48 / 66
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) = = 3 25 3 2 4 tangent is 3x + 4y = 25. an, as in Solution 1, an equation of the MAT 1001 Calculus I 49 / 66
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
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
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 = 2 3 32 3 2 2 3 = 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
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
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) ) = 1 1 + x 2. MAT 1001 Calculus I 54 / 66
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
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
Derivatives of Logarithmic Functions Example 25 Differentiate y = ln(x 3 + 1). Solution. To use the Chain Rule we let u = x 3 + 1. Then y = ln u, so y x = y u u x = 1 u u x = 1 x 2 + 1 (3x2 ) = 3x2 x 3 + 1 MAT 1001 Calculus I 57 / 66
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 10. 1 (2 + sin x) (2 + sin x) ln 10 x MAT 1001 Calculus I 58 / 66
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
Derivatives of Logarithmic Functions Logarithmic Differentiation Logarithmic Differentiation Example 28 Differentiate y = x3/4 x 2 + 1 (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 + 1 2 ln(x2 + 1) 5 ln(3x + 2) Differentiating implicitly with respect to x gives y x y = 3 4 1 x + 1 2 2x x 2 + 1 5 3 3x + 2 MAT 1001 Calculus I 60 / 66
Derivatives of Logarithmic Functions Logarithmic Differentiation Solution. Solving for y/x, we get y x = y y x y = 3 4x + x x 2 + 1 15 3x + 2 ( 3 4x + = x3/4 x 2 + 1 (3x + 2) 5 x x 2 + 1 15 ) 3x + 2 ( 3 4x + x x 2 + 1 15 3x + 2 ) MAT 1001 Calculus I 61 / 66
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
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
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
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
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 1 + 1 n. n n) MAT 1001 Calculus I 66 / 66