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?
On
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:
where the last step might be done by first defining $w = \sin(2x^2 - 17)$ and instead substituting $v = \tan(w)$.
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.