11.4 The Chain Rule.

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Presentation transcript:

11.4 The Chain Rule

Composite Functions Definition: A function m is a composite of functions f and g if m(x) = f [g(x)] The domain of m is the set of all numbers x such that x is in the domain of g and g(x) is in the domain of f.

Example 1 Given f(u) = 2u and g(x) = ex Find f [g(x)] and g [f(u)] f [g(x)] = f (ex) substitute g(x) for ex = 2(ex) plug ex into function f(u) = 2ex g [f(u)] = g (2u) substitute f(u) for 2u = e2u plug 2u into function g(x)

Example 2 Write each function as a composite of two simpler functions A) Y = 50 e-2x g(x) = -2x and f(u) = 50 eu B) and

C) Write y as a composite of two simpler functions y = (2x + 1)100 g(x) = 2x + 1 and f(u) = u 100 How do we find the derivative of a composite function such as the one above? What is the derivative of (2x + 1)100 Let’s investigate on the next slide!

Find the derivative of (2x + 1)2, (2x + 1)3, (2x + 1)4 and (2x + 1)100 Use the product rule: F(x) = (2x + 1)2 = (2x + 1) (2x + 1) F’(x) = (2)(2x+1) + (2)(2x+1) = 4(2x+1) F(x) = (2x + 1)3 = (2x + 1) (2x + 1)2 = (2x+1)(4x2 + 4x +1) F’(x) = (2)(4x2 + 4x +1) + (8x + 4)(2x + 1) = 2 (2x+1)2 + 4(2x+1)(2x+1) = 2 (2x+1)2 + 4(2x+1)2 = 6 (2x+1)2 Try (2x + 1)4 you should find out that f’(x) = 8 (2x+1)3 Look at the relationship between f(x) and f’(x), do you agree that the derivative of (2x + 1)100 is 200 (2x+1)99 ?

f(x) f’(x) (2x + 1)2 4(2x+1) = 2(2x+1) · 2 (2x + 1)3 Any composite function [g(x)] n n·[g(x)]n-1 · g’(x)

Generalized Power Rule What we have seen is an example of the generalized power rule: If u is a function of x, then

Chain Rule We have used the generalized power rule to find derivatives of composite functions of the form f (g(x)) where f (u) = un is a power function. But what if f is not a power function? It is a more general rule, the chain rule, that enables us to compute the derivatives of many composite functions of the form f(g(x)). Chain Rule: If y = f (u) and u = g(x) define the composite function y = m(x) = f [g(x)], then Or m’(x) = f’[g(x)] g’(x) provided that f’[g(x) and g’(x) exist

Example 3 A) h’(x) if h(x) = (5x + 2)3 h’(x) = 3 (5x + 2)2 (5x+2)’ Find the indicated derivatives A) h’(x) if h(x) = (5x + 2)3 h’(x) = 3 (5x + 2)2 (5x+2)’ = 3 (5x + 2)2 (5) =15 (5x + 2)2 B) y’ if y = (x4 – 5)5 y’= 5(x4 – 5)4 (x4 – 5)’ = 5(x4 – 5)4 (4x3) = 20x3 (x4 – 5)4

Example 3 Find the indicated derivatives C) D)

Example 4 A) y = u-5 and u = 2x3 + 4 Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x) A) y = u-5 and u = 2x3 + 4 The problem above can be solved by taking the derivative of y = (2x3 + 4) -5

Example 4 (continue) B) y = eu and u = 3x4 + 6 Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x) B) y = eu and u = 3x4 + 6 This problem can be solved by taking the derivative of

Example 5 C) y = ln u and u = x2 + 9x + 4 Find dy/du, du/dx, and dy/dx (express dy/dx as a function of x) C) y = ln u and u = x2 + 9x + 4 This problem can be solved by taking the derivative of y = ln (x2 + 9x + 4)

The chain rule can be extended to compositions of three or more functions. Given: Y = f(w), w = g(u), and u = h(x) Then

Example 6 y = h(x) = [ ln(1 + ex) ]3 find dy/dx Let y = w 3, w = ln u, and u = 1 + ex then

Example 6 (continue) y = h(x) = [ ln(1 + ex) ]3 find dy/dx Another way Derivative of the outside function The derivative of the function inside the [ ] Derivative of the function inside ( )

Examples for the Power Rule Chain rule terms are marked: Compare with the next slide

Examples for the Power Rule Chain rule terms are marked:

Examples for Exponential Derivatives Compare with the next slide

Examples for Exponential Derivatives Property of ln

Examples for Logarithmic Derivatives Compare with the next slide

Examples for Logarithmic Derivatives