Derivatives of Exponential and Logarithm Functions 10/17/2011 e0+h − e0 eh − 1 lim = lim = 1 h!0 h h!0 h The Derivative of y = ex Recall! ex is the unique exponential function whose slope at x = 0 is 1: m=1 The Derivative of y = ex Recall! ex is the unique exponential function whose slope at x = 0 is 1: m=1 e0+h − e0 eh − 1 lim = lim = 1 h!0 h h!0 h ex eh − 1 = lim h!0 h eh − 1 = ex lim h!0 h = ex ∗ 1 So d ex = ex dx d ex+h − ex ex = lim dx h!0 h The Derivative of y = ex ... eh − 1 lim = 1 h!0 h ex eh − 1 = lim h!0 h eh − 1 = ex lim h!0 h = ex ∗ 1 So d ex = ex dx The Derivative of y = ex ... eh − 1 lim = 1 h!0 h d ex+h − ex ex = lim dx h!0 h eh − 1 = ex lim h!0 h = ex ∗ 1 So d ex = ex dx The Derivative of y = ex ... eh − 1 lim = 1 h!0 h d ex+h − ex ex = lim dx h!0 h ex eh − 1 = lim h!0 h = ex ∗ 1 So d ex = ex dx The Derivative of y = ex ... eh − 1 lim = 1 h!0 h d ex+h − ex ex = lim dx h!0 h ex eh − 1 = lim h!0 h eh − 1 = ex lim h!0 h So d ex = ex dx The Derivative of y = ex ... eh − 1 lim = 1 h!0 h d ex+h − ex ex = lim dx h!0 h ex eh − 1 = lim h!0 h eh − 1 = ex lim h!0 h = ex ∗ 1 The Derivative of y = ex ... eh − 1 lim = 1 h!0 h d ex+h − ex ex = lim dx h!0 h ex eh − 1 = lim h!0 h eh − 1 = ex lim h!0 h = ex ∗ 1 So d ex = ex dx The Chain Rule Theorem Let u be a function of x. Then d du eu = eu : dx dx = 17e17x = cos(x)esin x p 2 = p2x+1 e x +1 2 x2+x Notice, every time: d ef (x) = f 0(x)ef (x) dx Examples Calculate... d 17x 1. dx e d sin x 2. dx e p d x2+x 3. dx e Notice, every time: d ef (x) = f 0(x)ef (x) dx Examples Calculate... d 17x 17x 1. dx e = 17e d sin x sin x 2. dx e = cos(x)e p p 2 2 3. d e x +x = p2x+1 e x +1 dx 2 x2+x Examples Calculate... d 17x 17x 1. dx e = 17e d sin x sin x 2. dx e = cos(x)e p p 2 2 3. d e x +x = p2x+1 e x +1 dx 2 x2+x Notice, every time: d ef (x) = f 0(x)ef (x) dx 1 1 = = eln(x) x Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 = dx ey d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! 1 1 = = eln(x) x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 = dx ey d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x 1 1 = = eln(x) x So dy 1 = dx ey d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx 1 1 = = eln(x) x d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 = dx ey 1 = x d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 1 = = dx ey eln(x) d 1 dx ln(x) = x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 1 1 = = = dx ey eln(x) x The Derivative of y = ln x To find the derivative of ln(x), use implicit differentiation! Remember: y = ln x =) ey = x Take a derivative of both sides of ey = x to get dy ey = 1 dx So dy 1 1 1 = = = dx ey eln(x) x d 1 dx ln(x) = x Does it make sense? d 1 dx ln(x) = x 1 f (x) = ln(x) 1 2 3 4 -1 3 2 1 f (x) = x 1 1 2 3 4 2x 2 = = x2 x 2x cos(x2) = sin(x2) Notice, every time: d f 0(x) ln(f (x)) = dx f (x) Examples Calculate d 2 1. dx ln x d 2 2. dx ln(sin(x )) Notice, every time: d f 0(x) ln(f (x)) = dx f (x) Examples Calculate 2x 2 1. d ln x2 = = dx x2 x 2x cos(x2) 2. d ln(sin(x2)) = dx sin(x2) Examples Calculate 2x 2 1. d ln x2 = = dx x2 x 2x cos(x2) 2. d ln(sin(x2)) = dx sin(x2) Notice, every time: d f 0(x) ln(f (x)) = dx f (x) d = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx (ln(2) is a constant!!!) For example: d 2x dx d 1 You try: log (x) = dx 2 ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a d = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx (ln(2) is a constant!!!) d 1 You try: log (x) = dx 2 ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d 2x dx = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x (ln(2) is a constant!!!) d 1 You try: log (x) = dx 2 ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d d 2x = ex ln(2) dx dx = ln(2) ∗ 2x d 1 You try: log (x) = dx 2 ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) dx dx (ln(2) is a constant!!!) d 1 You try: log (x) = dx 2 ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx (ln(2) is a constant!!!) 1 = ln(2) ∗ x The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx (ln(2) is a constant!!!) d You try: log (x) dx 2 The Calculus Standards: ex and ln x To get the other derivatives: ax = ex ln a ln x log x = a ln a For example: d d 2x = ex ln(2) = ln(2) ∗ ex ln(2) = ln(2) ∗ 2x dx dx (ln(2) is a constant!!!) d 1 You try: log (x) = dx 2 ln(2) ∗ x 1. If k = 1, then y = ex has this property and thus solves the equation. 2. For any k, y = ekx solves the equation too! d This equation, dx y = ky is an example of a differential equation. Differential equations Suppose y is some mystery function of x and satisfies the equation y 0 = ky Goal: What is y?? 2. For any k, y = ekx solves the equation too! d This equation, dx y = ky is an example of a differential equation. Differential equations Suppose y is some mystery function of x and satisfies the equation y 0 = ky Goal: What is y?? 1. If k = 1, then y = ex has this property and thus solves the equation. d This equation, dx y = ky is an example of a differential equation. Differential equations Suppose y is some mystery function of x and satisfies the equation y 0 = ky Goal: What is y?? 1. If k = 1, then y = ex has this property and thus solves the equation. 2. For any k, y = ekx solves the equation too! Differential equations Suppose y is some mystery function of x and satisfies the equation y 0 = ky Goal: What is y?? 1.