To prepare this question for differentiation depends on your view of the current function. As is, the function is ready for differentiation now with the product rule. It is possible to "simplify" it so that you can just use a single application of the chain rule, but it still gets...

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To prepare this question for differentiation depends on your view of the current function. As is, the function is ready for differentiation now with the product rule. It is possible to "simplify" it so that you can just use a single application of the chain rule, but it still gets to the same result.

Let `f(x)=x(4-x^2)^{1/2}`

With the product rule, the derivative is

`f'(x)=(4-x^2)^{1/2}+x(1/2)(4-x^2)^{-1/2}(-2x)` simplify right term

`=(4-x^2)^{1/2}-x^2/(4-x^2)^{1/2}` now get common denominator

`={4-x^2-x^2}/sqrt{4-x^2}`

`={2(2-x^2)}/sqrt{4-x^2}`

On the other hand, we can find the derivative by putting the x term into the square root before differentiation:

`f(x)=x(4-x^2)^{1/2}`

`=(4x^2-x^4)^{1/2}`

now differentiate

`f'(x)=1/2(4x^2-x^4)^{-1/2}(8x-4x^3)`

`={2x(2-x^2)}/sqrt{4x^2-x^4}`

`={2x(2-x^2)}/{xsqrt{4-x^2}}`

`={2(2-x^2)}/sqrt{4-x^2}`

**In both cases, there is a similar amount of algebra, but depending on how you feel about factoring, either one is easier to differentiate.**