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**Prove It** Like I say, it's always easier to break it into smaller parts.

$\displaystyle h = \frac{(2x^2 - 4x^6)^4}{2 + 2x^2}$

This is a quotient, so start with the quotient rule

$\displaystyle \frac{dh}{dx} = \frac{(2 + 2x)\frac{d}{dx}[(2x^2 - 4x^6)^4] - (2x^2 - 4x^6)^4\frac{d}{dx}(2 + 2x^2)}{(2 + 2x^2)^2}$

For the difficult looking derivative, use the chain rule.

$\displaystyle y = (2x^2 - 4x^6)^4$

Let $\displaystyle u = 2x^2 - 4x^6$ so that $\displaystyle y = u^4$

$\displaystyle \frac{du}{dx} = 4x - 24x^5, \frac{dy}{du} = 4u^3 = 4(2x^2 - 4x^6)^3$.

So $\displaystyle \frac{dy}{dx} = 4(4x - 24x^5)(2x^2 - 4x^6)^3$.

Therefore

$\displaystyle \frac{dh}{dx} = \frac{4(2 + 2x)(4x - 24x^5)(2x^2 - 4x^6)^3 - 4x(2x^2- 4x^6)^4}{(2 + 2x)^2}$.