`y = (1/2)(x^2)sqrt(16 - x^2)` Find the derivative of the function.

Textbook Question

Chapter 2, 2.4 - Problem 26 - Calculus of a Single Variable (10th Edition, Ron Larson).
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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You need to use the product and chain rules to evaluate the derivative of the function, such that:

`f'(x) = (1/2)(x^2)'(sqrt(16 - x^2)) + (1/2)(x^2)(sqrt(16 - x^2))'`

`f'(x) = x(sqrt(16 - x^2)) + (x^2)((-2x)/(4sqrt(1 - x^2)))`

`f'(x) = x(sqrt(16 - x^2)) - (x^3)/(2sqrt(1 - x^2))`

Hence, evaluating the derivative of the function, using the product rule, yields `f'(x) = x(sqrt(16 - x^2)) - (x^3)/(2sqrt(1 - x^2)).`

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