1. Chain Rule Derivative Question

Find the derivative of $y = (tan(8x))^2$.

Here is what I did, but apparently it's incorrect:

$y' = \frac {dy}{du} \frac{du}{dx} = [\frac{d}{du}u^2][\frac{d}{dx} tan 8x]$

$=2[tan 8x][8][\sec ^2 x]$
$=16[tan 8x][\sec ^2 x]$

2. Originally Posted by joatmon
Find the derivative of $y = (tan(8x))^2$.

Here is what I did, but apparently it's incorrect:

$y' = \frac {dy}{du} \frac{du}{dx} = [\frac{d}{du}u^2][\frac{d}{dx} tan 8x]$

$=2[tan 8x][8][\sec ^2 x]$
$=16[tan 8x][\sec ^2 x]$
It should be $\sec^2(8x).$

3. Make $\displaystyle u=\tan 8x \implies \frac{du}{dx} = 8\sec^28x$

4. $u=8x$

$w=tanu$

$y=w^2$

$\displaystyle\frac{dy}{dx}=\frac{dy}{dw}\frac{dw}{ du}\frac{du}{dx}$