Differentiate $\displaystyle tan^32x$ with respect to $\displaystyle x$

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- Sep 15th 2010, 04:03 AMPunchdifferentiate tan^3[2x]
Differentiate $\displaystyle tan^32x$ with respect to $\displaystyle x$

- Sep 15th 2010, 04:07 AMProve It
$\displaystyle y = \tan^3{2x}$.

Let $\displaystyle u = \tan{2x}$ so that $\displaystyle y = u^3$.

$\displaystyle \frac{du}{dx} = 2\sec^2{2x}$.

$\displaystyle \frac{dy}{du} = 3u^2 = 3\tan^2{2x}$.

Therefore $\displaystyle \frac{dy}{dx} = 6\sec^2{2x}\tan^2{2x}$.

You could use the Pythagorean Identity to clean this up further if you wanted...