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2-1 The Derivative and the Tangent Line Problem 2-2 Basic Differentiation Rules and Rates of Change 2-3 Product/Quotient Rule and Higher-Order Derivatives.

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Presentation on theme: "2-1 The Derivative and the Tangent Line Problem 2-2 Basic Differentiation Rules and Rates of Change 2-3 Product/Quotient Rule and Higher-Order Derivatives."— Presentation transcript:

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2 2-1 The Derivative and the Tangent Line Problem 2-2 Basic Differentiation Rules and Rates of Change 2-3 Product/Quotient Rule and Higher-Order Derivatives 2-4 Chain Rule 2-5 Implicit Differentiation 2-6 Related Rates (SKIP for now) Chapter 2 Differentiation

3 Alternate method: Rewrite y as a composite function: * This is called the Chain Rule.

4 Chain Rule: 2.4 Chain Rule If is the composite of and, then:

5 Differentiate h(x) = (x + 2) 4. We could find the derivative by expanding (x + 2) 4 and then using the Power Rule. h(x) = x 4 + 8x 3 + 24x 2 + 32x + 16 h’(x) = 4x 3 + 24x 2 + 48x + 32 This is a perfectly acceptable way to find the derivative, but multiplying out the binomial can be time consuming. So, now let’s find the derivative using the Chain Rule. Example 1

6 Using the Chain Rule, it will be helpful to identify the outside and inside before beginning. h(x) = (x + 2) 4 The outside = ( ) 4 and the inside = x + 2. So u = g(x) = x + 2 and f(u) = u 4. h(x) = f(g(x)) = (x + 2) 4. h’(x) = 4u 3 * 1 = 4(x + 2) 3 If is the composite of and, then:

7 Example 2: Find the derivative of. First rewrite the function with a rational exponent: The outside function = ( ) 1/3 and the inside function = 3x 2 – 1. This time, we are not able to FOIL, so the Chain Rule MUST be used. So u = g(x) = 3x 2 – 1 and f(u) = u 1/3. h(x) = f(g(x)) = (3x 2 – 1) 1/3.

8 Example 3: Find the derivative of. This could be done by the quotient rule, but the numerator is a constant and does not contain a variable so we could just rewrite h(x) and use the Chain Rule. The outside function = 2( ) -2 and the inside function = 4x – 1. So u = g(x) = 4x – 1 and f(u) = 2u -2. h(x) = f(g(x)) = 2(4x – 1) -2.

9 Example 4

10 You Try… Compute f ’(x).

11 Solutions

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13 Sub in for u

14 “Outside-inside” Rule It sometimes helps to think about the Chain Rule in terms of an outside function and inside function. In words, Differentiate the “outside” function f and leave “inside” function g(x) alone; then multiply by the derivative of the “inside” function. leave inside alone derivative of outside derivative of inside

15 Here is a faster way to find the derivative: Differentiate the outside function, leaving the inside function alone... …then multiply by the derivative of the inside function

16 Example 5 Find the derivative of. outside function = ( ) 3 inside function = x 2 + 3 The chain rule says take the derivative of the outside function leaving the inside function unchanged and then multiply by the derivative of the inside function. The derivative of the inside using the Power Rule The derivative of the outside leaving the inside unchanged

17 inside outside derivative of outside multiply by derivative of inside Example 6 leave inside

18 inside outside derivative of outside multiply by derivative of inside leave inside Example 7

19 Example 8 inside outside derivative of outside multiply by derivative of inside leave inside

20 1. y = (3x 2 + 2x + 1) 8 2. g(x) = cos (x – 1) 3. f(x) = sin(x² + x) 4. g’(x) = -sin(x – 1)(1)= -sin (x – 1) You Try… Differentiate each with respect to the indicated variable. f ’(x) = (2x + 1)cos(x 2 + 1)

21 Repeated Use of the Chain Rule Sometimes, we have to apply the chain rule more than once to calculate a derivative. Example 9 First rewrite the radical as an exponential. The derivative of the outside is simply the Power Rule, but the derivative of the inside would be found using the Chain Rule. (Chain Rule)

22 Continuing Now to simplify using our algebra skills. Rationalizing the denominator and Reducing.

23 Find the derivative of g(t) = tan (5 – sin 2t). Example 10

24 The chain rule can be used multiple times. (That’s what makes the “chain” in the “chain rule”!) Example 11

25 Example 12 Be careful not to confuse this with a product of sinx and tan 3x, therefore, Product Rule is not used. It is a composition of tan3x inside the function sinx, therefore Chain Rule is used. inside outer most function inside outside

26 You Try… Find.

27 You Try… 4. 5. Differentiate.

28 Be Careful…

29 Closure Explain how to apply the Chain Rule when there is more than 1 “inside” function like in the example:

30 Chain Rule and Product Rule Ex 1.

31 Ex 2. Differentiate y = (2x + 1) 5 (x 3 – x + 1) 4. GCF

32 Chain Rule and Quotient Rule Ex 3.

33 Ex 4. Differentiate.

34 Copyright © 2006 Brooks/Cole, a division of Thomson Learning, Inc. Chain Rule & Higher Derivatives Ex 5. Givenfind

35 Ex 6. Find the second derivative of. Always simplify the derivative before going on to the next derivative.

36 You Try…

37 Solution 4: Differentiate: The outside is, the inside is. First rewrite the square root as the exponent 1/2. To find the derivative we will need to use the Chain Rule and when we multiply by the derivative of the inside we will need to use the Quotient Rule to find the derivative of.

38 Simplifying more …

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40 Ex 7: Suppose F(x) = g(h(x)). If h (2) = 7, h’(2) = 3, g(2) = 9, g’(2) = 4, g(7) = 5, g’(7) = 11, find F’(2). F’(x) = g’(h(x))h’(x) F’(2) = g’(h(2))h’(2) = g’(7)h’(2) = (11)(3) = 33

41 Ex 8: Suppose g(x) = f(x 2 + 3(x – 1) + 5) and f '(6) = 21. Find g’(1). g’(x) = f’(x 2 + 3(x – 1) + 5) (x 2 + 3) g’(1) = f’(1 2 + 3(1 – 1) + 5) (1 2 + 3) = f ’(6)(4) = (21)(4) = 84

42 Ex 9: If f(x) = (7x 2 – 2x + 1) 3, find f ”(x). f’(x) = 3(7x 2 – 2x + 1) 2 (14x – 2) f”(x) = 3(7x 2 – 2x + 1) 2 (14) + (14x – 2) 3  2(7x 2 – 2x + 1)(14x – 2) = 52(7x 2 – 2x + 1) 2 + 6(7x 2 – 2x + 1)(14x – 2) 2

43 Closure Explain how to apply the Chain Rule when there is a Product or Quotient Rule involved.


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