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? 

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.
