# 6. Differentiating the exponential and logarithm functions

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1 1 6. Differentiating te exponential and logaritm functions We wis to find and use derivatives for functions of te form f(x) = a x, were a is a constant. By far te most convenient suc function for tis purpose is te exponential function wit base te special number e. Definition. Te number e, wic is approximately , is te number suc tat lim e = 1 Te number e is called Euler's number, after te great matematician Leonard Euler ( ). Te major reason for te use of e is te following teorem, wic says tat e x is its own derivative: Teorem. Dx [e x ] = e x. Proof. For any function f(x), recall tat f '(x) = lim f(x+) - f(x) Hence Dx [e x ] = lim e (x+) - e x = lim e x e - e x = lim e x (e - 1) = e x (1) [by te definition of e] = e x. We often call te function f(x) = e x te exponential function, since it involves taking an exponent. Example. Find Dx [5 e x ] Answer: 5 e x Derivatives get substantially more complicated wen we remember te cain rule:

2 2 Recall tat te composition g f of te functions f and g is te function (g f)(x) = g(f(x)). Tis means, "do te function f to x, ten do g to te result." Cain Rule. Suppose f and g ave derivatives. Ten (g f)'(x) = g'(f(x)) f '(x). Tis means tat we can use te cain rule for te exponential function as well: Teorem. (Exponential cain rule) Dx [e f(x) ] = e f(x) Dx[f(x)]. Proof. Tis is te cain rule were g(y) = e y. Example. Find Dx [e 6x ] Dx [e 6x ] = e 6x Dx [6x ] = 6 e 6x Example. Find Dx [e x2 ] Dx [e x2 ] = e x2 Dx [x 2 ] = 2x e x2 Example. Find Dx [5 e 3x ] Dx [5 e 3x ] = 5 e 3x Dx [3x ] = 15 e 3x Example. Find Dx [2x e 4x ] Dx [2x e 4x ] = (2x) Dx [ e 4x ] + e 4x Dx [2x ] [by te Product Rule] = (2x) e 4x Dx[4x] + e 4x 2 = (2x) e 4x (4) + 2 e 4x = 8x e 4x + 2 e 4x Example. If f(t) = e 4t e 2t find f '(t). Use te quotient rule (1 + e 2t ) Dt[e 4t ] - e 4t Dt[1 + e 2t ] f '(t) = (1 + e 2t ) 4 e 4t - e 4t 2 e 2t = e 4t + 4 e 2t e 4t -2 e 4t e 2t =

3 3 4 e 4t + 4 e 6t -2 e 6t = e 4t + 2 e 6t = Closely related to te exponential function is te logaritm function. Recall te natural logaritm of x, written ln(x), is te number suc tat e ln(x) = x. Tus ln(x) is te power to wic e is raised to yield te number x. It makes sense only if x>0. Teorem. Dx [ln(x)] = 1/x if x > 0. Wy is it true tat Dx [ln(x)] = 1/x if x > 0? Te argument is a clever one. Let g(x) = e [ln(x)]. Ten by te Exponential Cain Rule we ave Dx[g(x)] = e [ln(x)] Dx [ln(x)] But g(x) = e [ln(x)] = x, so Dx[g(x)] = 1. Hence 1= e [ln(x)] Dx [ln(x)] 1= x Dx [ln(x)] Dx [ln(x)] = 1/x Example. If g(x) = 3 ln(x) find g '(2). g '(x) = 3 Dx [ln(x)] = 3 /x. Hence g '(2) = 3/2 = 1.5. Teorem. (Logaritm cain rule) Dx [ln (f(x))] Dx [f(x)] f '(x) = = f(x) f(x) Proof. Write g(x) = ln(x). Ten Dx [ln (f(x))] = Dx [g (f(x))] = g '(f(x)) f '(x) [by te Cain Rule] = [1 / f(x)] f '(x) = f '(x) / f(x) Example. Find Dx [ln ()] Dx [ln ()]

4 4 = Dx [] = 2x Example. Find f '(x) if f(x) = 2x ln () f '(x) = Dx [2 x ln ()] = (2x) Dx [ ln ()] +[ ln ()] Dx [2 x] = (2x) Dx [ ] [ ln ()] (2) = (2x) (2x) ln () = 4x ln () Example. Find f '(x) if f(x) = ln() x+1 By te quotient rule we ave f '(x) = (x+1) Dx [ln()] - [ln()] Dx [x+1] (x+1) 2 f '(x) = (x+1) [1/()] Dx [] - [ln()] (1) (x+1) 2 f '(x) = (x+1) [1/()] (2x) - ln() (x+1) 2 f '(x) = (x+1)(2x) /() - ln() (x+1) 2 Simplify by multiplying numerator and denominator by (x 2 +1).

5 5 f '(x) = (x+1)(2x) - () ln() (x+1) 2 () Example. Find f '(x) if f(x) = e 3x ln(2x + 1). Use te product rule: f ' (x) = e 3x Dx[ln(2x + 1)] + ln(2x + 1) Dx[ e 3x ] f ' (x) = e 3x [1/(2x + 1)]Dx[2x+1] + ln(2x + 1) e 3x (3) f ' (x) = e 3x [2/(2x + 1)]+ 3 e 3x ln(2x + 1) f ' (x) = 2 e 3x /(2x + 1)+ 3 e 3x ln(2x + 1) We sall often ave to solve equations involving exponents or logaritms. Tese will arise wen we need to find critical numbers, or were a function is increasing of decreasing, or finding maxima or minima. Tese typically involve te identities ln (e c ) = c e ln(b) = b Example. Solve for x if e 2x = 5. Give your answer (a) exactly (b) to 5 decimal places e 2x = 5 ln[e 2x ] = ln(5) 2 x = ln(5) x = [ln(5)]/2 Hence (a) as answer [ln(5)]/2. To find te answer to (b), use your calculator: to 5 decimal places means Example. Solve for x if ln(3x) - 4 = 0. Give your answer (a) exactly (b) to 5 decimal places Solution ln(3x) - 4 = 0 ln(3x) = 4 e [ln(3x)] = e 4 3x = e 4 (a) x = (e 4 )/3 (b) x = Example. Solve for x if e 2x - 5 x e 2x = 0. e 2x - 5 x e 2x = 0 Factor: e 2x (1-5 x) = 0. If a product is 0, one of te factors is 0. Hence eiter e 2x = 0 or 1-5x = 0 But e 2x > 0 since it is a positive number raised to a power, so it is never 0. Hence 1-5x = 0

6 6 5x = 1 x = 1/5 Example. Solve for x if x 2 e 3x + 12 e 3x = 7 x e 3x Rewrite as an expression = 0: x 2 e 3x - 7 x e 3x + 12 e 3x = 0 Factor e 3x (x 2-7 x + 12 ) = 0 Since e 3x >0, we must ave x 2-7 x + 12 = 0 (x-4)(x-3) = 0 x - 4 = 0 or x-3 = 0 x = 4 or x = 3 Example. Solve for x if 3 e -2x - 5 e -6x = 0. Give te answer to 4 decimal places. 3 e -2x - 5 e -6x = 0. 3 e -2x = 5 e -6x 3 e -2x / e -6x = 5 3 e -2x -(-6x) = 5 3 e 4x = 5 e 4x = 5/3 ln e 4x = ln(5/3) 4x = ln(5/3) x = (1/4) ln(5/3) = We can use tese new functions to answer questions about relative extrema and absolute extrema. Example. Let f(x) = x e -2x. (a) Tell were f is increasing, and were f is decreasing. (b) Locate and classify all critical numbers. (c) Find te absolute maximum and absolute minimum for 0 x 4, and were tey occur. f '(x) = x Dx [e -2x ] + e -2x Dx[x] = x e -2x Dx [-2x] + e -2x (1) = x e -2x (-2) + e -2x = -2x e -2x + e -2x To find te critical numbers we solve f '(x) = 0: -2x e -2x + e -2x = 0 Multiply by e 2x : -2x + 1 = 0 2x = 1 x = 1/2 We now find te cart for f '(x) by finding te value at points not equal to a critical number:

7 7 (+) 1/2 (-) (a) f is increasing for x < 1/2. f is decreasing for x > 1/2. (b) Te only critical number is 1/2 relative maximum (c) For 0 x 4 we form a cart including te endpoints and critical numbers in te interval x f(x) 1/ Te absolute maximum is at x = 1/2. Te absolute minimum is 0 at x = 0. Example. Let f(x) = ln (x) for x > 0. 2x (a) Tell were f is increasing, and were f is decreasing. (b) Locate and classify all critical numbers. (c) Find te absolute maximum and absolute minimum for 1 x 10, and were tey occur. Give answers to 3 decimal places We must solve f '(x) = 0. By te quotient rule 2x Dx[ ln(x)] - (ln (x)) Dx[2x] f ' (x) = (2x) 2 2x [1/x] - (ln (x)) (2) f ' (x) = x ln (x) f ' (x) = x 2 Hence f '(x) = 0 wen 2-2 ln(x) = 0 2 ln(x) = 2 ln(x) = 1 e ln(x) = e 1 x = e We find te cart for f '(x) as follows, remembering tat x>0. Note f(0) is meaningless. 0 (+) e (-) Hence (a) f is increasing for 0<x<e and f is decreasing for x>3. (b) Te only critical number is x = e, wic is a relative maximum.

8 8 (c) We must consider te endpoints x = 1 and x = 10 togeter wit critical numbers in te interval. Since 1<e<10, we must consider x= e. Hence our cart of values is x f(x) 1 ln (1) / (2(1)) = 0 10 ln (10) / 20 = e ln (e) / (2e) =.184 Te absolute maximum is.184 at x = e = Te absolute minimum is 0 at x = 1. Example. Te concentration C(t) of a drug in te blood t ours after a pill is swallowed is given by C(t) = 7 e -0.2t - 7 e -0.5t mg/l. (a) Wen is te concentration at a maximum? (b) Wat is te maximum concentration? (c) For wic t is te concentration increasing? (d) For wic t is te concentration decreasing? Give answers to 2 decimal places. We differentiate C(t): C '(t) = 7 e -0.2t (-0.2) - 7 e -0.5t (-0.5) = -1.4 e -0.2t +3.5 e -0.5t We need te critical numbers, so we solve C '(t) = e -0.2t +3.5 e -0.5t = 0 Divide by e -0.2t : e -0.5t / e -0.2t = e -0.5t+0.2t = e -0.3t = e -0.3t = 1.4 e -0.3t = 1.4/3.5 = 0.4 ln ( e -0.3t ) = ln( 0.4) -0.3 t = ln (0.4) t = (ln (0.4))/(-0.3) t = 3.05 ours Note te cart for C ' is ten (+) 3.05 (-) Hence te answers are (a) t = 3.05 ours. (b) C(3.05) = 2.28 mg/l (c) 0<t< 3.05 ours (d) t > 3.05 ours

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