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§2.3 The Chain Rule and Higher Order Derivatives

§2.3 The Chain Rule and Higher Order Derivatives. The student will learn about. composite functions,. the chain rule, and. nondifferentiable functions. Composite Functions. Definition. A function m is a composite of functions f and g if

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§2.3 The Chain Rule and Higher Order Derivatives

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  1. §2.3 The Chain Rule and Higher Order Derivatives The student will learn about composite functions, the chain rule, and nondifferentiable functions.

  2. Composite Functions Definition. A function m is a composite of functions f and g if m (x) = f ◦ g = f [ g (x)] This means that x is substituted into g first. The result of that substitution is then substituted into the function f for your final answer.

  3. ExamplesLet f (u) = u 3 , g (x) = 2x + 5, and m (v) = │v│. Find: f [ g (x)] = f (2x + 5) = (2x + 5)3 g [ f (x)] = g (x3) = 2x 3 + 5 m [ g (x)] = m (2x + 5) = │2x + 5│

  4. Chain Rule: Power Rule. We have already made extensive use of the power rule with xn, We wish to generalize this rule to cover [u (x)]n, where u (x) is a composite function. That is it is fairly complicated. It is not just x.

  5. Chain Rule: Power Rule. That is, we already know how to find the derivative of f (x) = x 5 We now want to find the derivative of f (x) = (3x 2 + 2x + 1) 5 What do you think that might be?

  6. Chain Rule: Power Rule. General Power Rule. [Chain Rule] If u (x) is a function, n is any real number, and If f (x) = [u (x)]n then f ’ (x) = n un – 1 u’ or The chain * * * * * VERY IMPORTANT * * * * *

  7. Example Find the derivative of y = (x3 + 2) 5. Chain Rule Let the ugly function be u (x) = x3 + 2. Then 5 (x3 + 2) 4 3x2 = 15x2(x3 + 2)4

  8. Examples Find the derivative of: y = (x+ 3) 2 y’ = 2 (x+ 3)(1) = 2 (x + 3) y = (4 – 2x 5) 7 y’ = 7 (4 – 2x 5) 6(- 10x 4) y’ = - 70x 4 (4 – 2x 5) 6 y = 2(x3 + 3) – 4 y’ = - 8(x3 + 3) – 5(3x 2) y’ = - 24x 2 (x3 + 3) – 5

  9. Example Find the derivative of y = Rewrite as y = (x 3 + 3) 1/2 Then y’ = 1/2 Then y’ = 1/2 (x 3 + 3) – 1/2 Then y’ = 1/2 (x 3 + 3) – 1/2(3x2) = 3/2 x2 (x3 + 3) –1/2 Try y = (3x 2 - 7) - 3/2 y’ = (- 3/2) (3x 2 - 7) - 5/2(6x) = (- 9x) (3x 2 - 7) - 5/2

  10. Example Find f ’ (x) if f (x) = We will use a combination of the quotient rule and the chain rule. Let the top be t (x) = x4, then t ‘ (x) = 4x3 Let the bottom be b (x) = (3x – 8)2, then using the chain rule b ‘ (x) = 2 (3x – 8) 3 = 6 (3x – 8)

  11. Remember Def: The instantaneous rate of change for a function, y = f (x), at x = a is: This is the derivative. Sometimes this limit does not exist. When that occurs the function is said to be nondifferentiable.

  12. Remember Def: The instantaneous rate of change for a function, y = f (x), at x = a is: This is the derivative and a graphing way to represent the derivative is as the slope of the curve. This means that at some points on some curves the slope is not defined.

  13. “Corner point”

  14. Vertical Tangent

  15. Discontinuous Function

  16. "If a function f …"

  17. Summary. If y = f (x) = [u (x)]n then Nondifferentiable functions.

  18. ASSIGNMENT §2.3 on my website 11, 12, 13.

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