Lecture 11 - Derivatives of General Exponential and Inverse Functions

Calculus I - Lecture 11 - Derivatives of General Exponential and Inverse Functions

Lecture Notes: http://www.math.ksu.edu/˜gerald/math220d/ Course Syllabus: http://www.math.ksu.edu/math220/spring-2014/indexs14.html

Gerald Hoehn (based on notes by T. Cochran)

February 26, 2014 Derivative of General Exponential functions We know: d ex = ex dx By the chain rule: d f (x) x e = e f 0(x) dx · Example: Find the derivative of 2x . Solution: Recall 2 = eln 2, so 2x = eln 2 x = e(ln 2)x . d d d Thus 2x = e(ln 2)x = e(ln 2)x ((ln 2)x) dx dx dx = (ln 2) 2x . · Theorem d bx = (ln b) bx , for any base b > 0. dx · d 2 Example: Find 7x . dx Solution: We apply the chain rule with outer function f (u) = 7u and inner function g(x) = x2: d 2 2 d 7x = (ln 7) 7(x ) x2 dx · · dx 2 = 2 ln 7 x 7x · · d x Example: Find 55 . dx Solution:

d x x d 55 = (ln 5) 55 5x dx · · dx x = (ln 5) 55 (ln 5) 5x · · · x = ln2 5 5x 55 · · The Derivative of the Natural Logarithm function

dy Example: Let y = ln x. Find . dx Solution: We have ey = eln x = x. We take now the derivative on both sides: d d ey = x dx dx dy ey = 1 (by the chain rule) · dx dy 1 1 Thus: = = , since ey = x. dx ey x We have shown the following rule: Theorem d 1 ln x = dx x d Example: Find ln(x 5x3). dx − Solution: d 1 d ln(x 5x3) = (x 5x3) dx − x 5x3 · dx − 1 − = (1 15x2) x 5x3 · − − 1 15x2 = − x 5x3 − d Example: Find x ln x. dx Solution: d d d x ln x = (x) ln x + x (ln x) dx dx · · dx 1 = 1 ln x + x · · x = 1 + ln x d Example: Compute log x for any base b > 0. dx b Solution: ln x We have log x = . b ln b Thus: d d ln x 1 d 1 1 log x = = ln x = . (Remember) dx b dx ln b ln b · dx ln b · x d Example: Compute log (log (x)). dx 5 5 Solution: d 1 1 d log5(log5(x)) = log5 x dx ln 5 · log5 x · dx 1 1 1 1 = ln 5 · log5 x · ln 5 · x 1 = ln2 5 x log x · 5 Logarithmic Diﬀerentiation dy Let y = f (x), y 0 = dx = f 0(x). Theorem d y ln y = 0 dx y Indeed, by the chain rule: d d 1 d 1 y 0 ln y = ln(f (x)) = f (x) = f 0(x) = . dx dx f (x) · dx f (x) · y This is sometimes helpful to compute the derivative of a function which is mainly a combination of products, quotients or powers: 1. Take‘ ln’ of both sides and expand using the folowing rules:

I ln(AB) = ln A + ln B I ln(A/B) = ln A ln B n − I ln(A ) = n ln A d 2. Take‘ dx ’ of both sides, using the theorem for the left side. 3. Solve for y’. dy Example: Find dx by logarithmic diﬀerentiation, given y = √4 x sin3 x. Solution: Step 1: ln y = ln √4 x sin3 x 4 3 = ln(√x) + ln(sin x) = ln(x1/4) + ln((sin x)3) 1 = 4 ln x + 3 ln(sin x) Step 2: d d 1 d ln y = ln x) + (3 ln(sin x)) dx dx 4 dx y 1 1 1 0 = + 3 cos x y 4 · x · sin x · Step 3: 1 y 0 = + 3 cot x y 4x ·

1 4 3 y 0 = + 3 cot x √x sin x 4x Derivative of Inverse Functions The trick we have used to compute the derivative of the natural logarithm works in general for inverse functions. 1 Recall that the inverse function f − (x) of a function f (x) is 1 defined by the property that f (f − (x)) = x. 1 Warning: Do not confuse f − (x) with the reciprocal 1/f (x). Theorem d 1 1 f − (x) = 1 dx f 0(f − (x)) 1 Proof: We differentiate both sides of f (f − (x)) = x: d 1 d f (f − (x)) = x dx dx 1 d 1 f (f (x)) f − (x) = 1 (by the chain rule) 0 − · dx d 1 1 Thus: f − (x) = 1 . dx f 0(f − (x)) Graphical Understanding Inverses of the trigonometric functions Inverses of the trigonometric functions d Example: Find arcsin x. dx Solution: 1 arcsin x = sin− x is the inverse function of sin x. d Since sin x = cos x, by the theorem for inverse functions: dx d d 1 1 arcsin x = sin− x = 1 dx dx cos(sin− x) Using sin2 z + cos2 z = 1 or cos z = 1 sin2 z, we obtain: − d 1 1 arcsin x = =p dx 2 1 √1 x2 1 sin (sin− x) − − q The arcus tangent function y = arctan x is defined for x R with ∈ π < y < π . − 2 2

d 1 Example: Find arctan x = tan− (x). dx Solution: We use again the theorem for the derivative of inverse functions. d 1 Since tan x = sec2 x = we get: dx cos2 x d d 1 1 2 1 arctan x = tan− x = 2 1 = cos (tan− x). dx dx sec (tan− x) 1 Using sin2 z + cos2 z = 1 or tan2 z + 1 = , we obtain: cos2 z d 1 1 arctan x = 2 1 = 2 dx 1 + tan (tan− x) 1 + x Theorem

f (x) f 0(x) 1 arcsin x √1 x2 − 1 arccos x − √1 x2 − 1 arctan x 1 + x2 1 arcsecx x √x2 1 | | −