Differentiate y=Cot²(sinx) $$ y = \cot^2(\sin x) $$
How do I differentiate that? I tried using chain rule but I don't understand how to differentiate $\cot^2(\sin x)$.
 A: Try to see this as $$[h(\ g\ (f(x))\ )]' = h'(g(f(x)))[g(f(x))]' = h'(g(f(x)))\ g'(f(x))\ f'(x) $$
Then you will have
$$\frac{d}{dx}\cot^2(\sin x) = -2\cot(\sin x)\  \csc^2(\sin x)  \cos x $$
Where $\frac{d}{dx}(\cot x) = -\csc^2 x$ and $\frac{d}{dx}(\sin x) = \cos x$. 
A: Hint : Use chain rule.
$\frac{d}{dx}(g(f(x)))= g'(f(x)).f'(x)$
A: by the chaine rule we obtain $2\,\cot \left( \sin \left( x \right)  \right)  \left( -1- \left( \cot
 \left( \sin \left( x \right)  \right)  \right) ^{2} \right) \cos
 \left( x \right) 
$
$(\cot(x))'=-1-\cot(x)^2$
A: First note that
the chain rule is
$$ 
\frac{d}{dx}f(g(x))=\frac{d}{dg(x)}f(g(x))\frac{d}{dx}g(x)
$$
This implies that
$$
\frac{d}{dx}(f(x))^2=2f(x)\frac{d}{dx}f(x)
$$
$$
\frac{d}{dx}\cot(f(x)) = -\csc^2(f(x))\frac{d}{dx}f(x)
$$
Also note that
$$
\frac{d}{dx}\sin x=\cos x
$$
So now we have
$$
\frac{d}{dx}\left[\cot^2(\sin x)\right]= 2 \cot(\sin x) \frac{d}{dx}\left[\cot(\sin x)\right] $$
$$= -2\cot(\sin x) \csc^2(\sin x)\frac{d}{dx}\left[\sin x\right]= -2 \cot(\sin x) \csc^2(\sin x)\cos x
$$
