Derivatives of Exponential and Logarithm Functions 10/17/2011 The - PowerPoint PPT Presentation

Derivatives of Exponential and Logarithm Functions 10/17/2011

The Derivative of y = e x Recall! e x is the unique exponential function whose slope at x = 0 is 1: m=1

The Derivative of y = e x Recall! e x is the unique exponential function whose slope at x = 0 is 1: m=1 e 0+ h − e 0 e h − 1 lim = lim = 1 h h h → 0 h → 0

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0 e x + h − e x d dx e x = lim h h → 0

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0 e x + h − e x d dx e x = lim h h → 0 e x � e h − 1 � = lim h h → 0

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0 e x + h − e x d dx e x = lim h h → 0 e x � e h − 1 � = lim h h → 0 e h − 1 = e x lim h h → 0

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0 e x + h − e x d dx e x = lim h h → 0 e x � e h − 1 � = lim h h → 0 e h − 1 = e x lim h h → 0 = e x ∗ 1

The Derivative of y = e x ... e h − 1 lim = 1 h h → 0 e x + h − e x d dx e x = lim h h → 0 e x � e h − 1 � = lim h h → 0 e h − 1 = e x lim h h → 0 = e x ∗ 1 So d dx e x = e x

The Chain Rule Theorem Let u be a function of x. Then dx e u = e u du d dx .

Examples Calculate... dx e 17 x d 1. dx e sin x d 2. √ x 2 + x d 3. dx e

Examples Calculate... dx e 17 x = 17 e 17 x d 1. dx e sin x = cos( x ) e sin x d 2. √ √ x 2 + x = 2 x +1 x 2 +1 d 3. dx e x 2 + x e √ 2

Examples Calculate... dx e 17 x = 17 e 17 x d 1. dx e sin x = cos( x ) e sin x d 2. √ √ x 2 + x = 2 x +1 x 2 +1 d 3. dx e x 2 + x e √ 2 Notice, every time: d dx e f ( x ) = f ′ ( x ) e f ( x )

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation!

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒ Take a derivative of both sides of e y = x to get dy dx e y = 1

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒ Take a derivative of both sides of e y = x to get dy dx e y = 1 So dx = 1 dy e y

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒ Take a derivative of both sides of e y = x to get dy dx e y = 1 So dx = 1 1 dy e y = e ln( x )

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒ Take a derivative of both sides of e y = x to get dy dx e y = 1 So dx = 1 e ln( x ) = 1 1 dy e y = x

The Derivative of y = ln x To find the derivative of ln( x ), use implicit differentiation! Remember: e y = x y = ln x = ⇒ Take a derivative of both sides of e y = x to get dy dx e y = 1 So dx = 1 e ln( x ) = 1 1 dy e y = x dx ln( x ) = 1 d x

Does it make sense? dx ln( x ) = 1 d x 1 f ( x ) = ln( x ) 1 2 3 4 -1 3 2 f ( x ) = 1 x 1 1 2 3 4

Examples Calculate dx ln x 2 d 1. dx ln(sin( x 2 )) d 2.

Examples Calculate dx ln x 2 = 2 x x 2 = 2 d 1. x dx ln(sin( x 2 )) = 2 x cos( x 2 ) d 2. sin( x 2 )

Examples Calculate dx ln x 2 = 2 x x 2 = 2 d 1. x dx ln(sin( x 2 )) = 2 x cos( x 2 ) d 2. sin( x 2 ) Notice, every time: dx ln( f ( x )) = f ′ ( x ) d f ( x )

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: d dx 2 x

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: dx 2 x = d d dx e x ln(2)

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: dx 2 x = d d dx e x ln(2) = ln(2) ∗ e x ln(2) (ln(2) is a constant!!!)

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: dx 2 x = d d dx e x ln(2) = ln(2) ∗ e x ln(2) = ln(2) ∗ 2 x (ln(2) is a constant!!!)

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: dx 2 x = d d dx e x ln(2) = ln(2) ∗ e x ln(2) = ln(2) ∗ 2 x (ln(2) is a constant!!!) You try: d dx log 2 ( x )

The Calculus Standards: e x and ln x To get the other derivatives: a x = e x ln a log a x = ln x ln a For example: dx 2 x = d d dx e x ln(2) = ln(2) ∗ e x ln(2) = ln(2) ∗ 2 x (ln(2) is a constant!!!) 1 You try: d dx log 2 ( x ) = ln(2) ∗ x

Differential equations Suppose y is some mystery function of x and satisfies the equation y ′ = ky Goal: What is y ??

Differential equations Suppose y is some mystery function of x and satisfies the equation y ′ = ky Goal: What is y ?? 1. If k = 1, then y = e x has this property and thus solves the equation.

Differential equations Suppose y is some mystery function of x and satisfies the equation y ′ = ky Goal: What is y ?? 1. If k = 1, then y = e x has this property and thus solves the equation. 2. For any k , y = e kx solves the equation too!

Differential equations Suppose y is some mystery function of x and satisfies the equation y ′ = ky Goal: What is y ?? 1. If k = 1, then y = e x has this property and thus solves the equation. 2. For any k , y = e kx solves the equation too! d This equation, dx y = ky is an example of a differential equation .