# differentiate tan^3[2x]

• Sep 15th 2010, 05:03 AM
Punch
differentiate tan^3[2x]
Differentiate $tan^32x$ with respect to $x$
• Sep 15th 2010, 05:07 AM
Prove It
$y = \tan^3{2x}$.

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

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

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

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

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