How is the chain method used in finding this derivative? 
Find the derivative of $\tan^3[\sin(2x^2-17)]$. 

Sorry if my question is a little too specific but I am confused on this trig equation. After completing the derivative I was wondering why does the $3tan^2$ not distribute to $sec^2$? Is there a rule for this? How come the exponents and the power of $2$ don't get placed onto $sec^2$? Am I missing out on some of the properties of the chain rule? 
 A: You're probably getting confused by trying to do too much at once. Rather than doing the whole calculation in a single step, apply them one at a time. For example, apply the power rule
$$ \mathrm{d}(u^n) = n u^{n-1} \mathrm{d}u $$
to get
$$ \mathrm{d}\left( \tan^3[\sin(2x^2-17)] \right)
= 3 \left( \tan[\sin(2x^2-17)] \right)^2 \mathrm{d}\left( \tan[\sin(2x^2-17)] \right)$$
If you have trouble even with that, then introduce a lot of new variables to hide the complexity of the formula.
E.g. if you define $v = \tan[\sin(2x^2-17)] $, then the question is to differentiate $v^3$. And you should do so by:


*

*Forget how $v$ is expressed in terms of $x$

*Compute the derivative of $v^3$ 

*Substitute back in how $v$ is in terms of of $x$


where the last step might be done by first defining $w = \sin(2x^2 - 17)$ and instead substituting $v = \tan(w)$.
A: Let $y=\tan^3[\sin(2x^2-17)]$ then
$$y=\bigg[\underbrace{\tan(\sin (2x^2-17))}_{\text{base}}\bigg]^3$$ and so applying the Power Rule, we get
$$\frac{dy}{dx}=3\cdot \bigg[\underbrace{\tan(\sin (2x^2-17))}_{\text{base}}\bigg]^2\cdot\frac{d}{dx}(\text{base}).$$ Now,
$$\begin{align}\frac{d}{dx}(base)&=\frac{d}{dx}(\tan(\underbrace{\sin (2x^2-17)}_{\text{angle}})\qquad\text{then apply Chain Rule to get}\\
&=\sec^2(\underbrace{\sin(2x^2-17)}_{\text{angle}})\cdot \frac{d}{dx}(\text{angle})
\end{align}$$
Lastly,
$$\begin{align}
\frac{d}{dx}(\text{angle})&=\frac{d}{dx}(\sin(2x^2-17))\qquad\text{then apply again the Chain Rule to get}\\
&=\cos(2x^2-17)\cdot \frac{d}{dx}(2x^2-17)\\
&=[\cos(2x^2-17)]\cdot 4x\\
&=4x\cos(2x^2-17).
\end{align}$$
Doing substitutions, we eventually get $\frac{dy}{dx}$.
Hope this helps.
