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11.4 The Chain Rule.

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Presentation on theme: "11.4 The Chain Rule."— Presentation transcript:

1 11.4 The Chain Rule

2 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.

3 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)

4 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

5 C) Write y as a composite of two simpler functions
y = (2x + 1)100 g(x) = 2x 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!

6 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) = (2x+1) (2x+1)(2x+1) = (2x+1) (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 ?

7 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)

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

9 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

10 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

11 Example 3 Find the indicated derivatives C) D)

12 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

13 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

14 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)

15 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

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

17 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 ( )

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

19 Examples for the Power Rule
Chain rule terms are marked:

20 Examples for Exponential Derivatives
Compare with the next slide

21 Examples for Exponential Derivatives
Property of ln

22 Examples for Logarithmic Derivatives
Compare with the next slide

23 Examples for Logarithmic Derivatives


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