I am trying to do this homework problem and I have no idea how to approach it. I have tried many methods, all resulting in failure. I went to the books website and it offers no help. I am trying to find the derivative of the function
$$y=\cot^2(\sin \theta)$$
I could be incorrect but a trig function squared would be the result of the trig function with the angle value and then squared. Not the angle value squared, that would give a different answer. Knowing this I also know that I can not use the table of simple trig derivatives so I know I can't just take the derivative as
$$y=\cot^2(x)$$
$$ x=\sin(\theta)$$
This does not help because I can't get the derivative of cot squared. What I did try to do was rewrite it as $\frac{\cos x}{\sin x}\frac{\cos x}{\sin x}$ and then find the derivative of that but something went wrong with that and it does not produce an answer that is like the one in the book. In fact the book gets a csc squared in the answer so I know they are doing something very different.
Best Answer
Indeed, $\cot^2(a)$ means $$\left(\cot (a)\right)^2.$$
You need to apply the Chain Rule twice: first, to deal with the square: set $g(u)=u^2$ as your "outside function", and $u=f(\theta) = \cot(\sin(\theta))$ as your inside function. Since $g'(u) = 2u$, then $$\frac{d}{d\theta}\cot^2(\sin(\theta)) = \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2 = g'(u)f'(\theta) = 2uf'(\theta) = 2\cot\bigl(\sin(\theta)\bigr)f'(\theta).$$ Now let's deal with $f'(\theta)$; we have $f(\theta) = \cot\bigl(\sin(\theta)\bigr)$. The "outside function" is $h(u) = \cot(u)$, the "inside function" is $u(\theta) = \sin(\theta)$. Since $h'(u) = -\csc^2(u)= -\left(\csc(u)\right)^2$, and $f'(\theta) = \cos(\theta)$, we have: $$\frac{d}{d\theta}\cot\bigl(\sin(\theta)\bigr) = h'(u)u'(\theta) = -\csc^2(\sin\theta)\cos(\theta).$$
Putting it all together: $$\begin{align*} \frac{d}{d\theta}\cot^2\bigl(\sin(\theta)\bigr) &= \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)^2\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \frac{d}{d\theta}\left(\cot\bigl(\sin(\theta)\bigr)\right)\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot \left(-\csc^2\left(\sin(\theta)\right)\left(\frac{d}{d\theta}\sin(\theta)\right)\right)\\ &= 2\left(\cot\bigl(\sin(\theta)\bigr)\right)\cdot\left(-\csc^2\left(\sin(\theta)\right)\cos(\theta)\right)\\ &= -2\cot\bigl(\sin(\theta)\bigr)\csc^2\bigl(\sin(\theta)\bigr)\cos(\theta). \end{align*}$$