`h(t) = (t + 1)^(2/3) (2t^2 - 1)^3` Find the derivative of the function.

Textbook Question

Chapter 3, 3.4 - Problem 19 - Calculus: Early Transcendentals (7th Edition, James Stewart).
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gsenviro | College Teacher | (Level 1) Educator Emeritus

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Using the product rule of derivative, we can solve this as:

`h'(t) = (2/3) (t+1)^(-1/3) *(1) * (2t^2-1)^3 `

`+ (t+1)^(2/3) * 3*(2t^2-1)^2 * (4t)`

`= (2t^2-1)^2 (t+1)^(-1/3) [ (2/3)(2t^2-1) + (t+1)*(12t)]`

`= (2t^2-1)^2(t+1)^(-1/3)(40/3 t^2 +12t-2/3)`

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