Derivative of $\csc(f(x))$ Could someone explain how can I find the derivative:
$$  \frac{d}{dx}\csc[f(x)] = ? $$
 A: Looking into the definition of $\csc{v} = \dfrac{1}{\sin(v)}$ and the chain rule of $f(g(x)) = g'(x)\cdot f'(g(x))$ (same, if there is more function nested functions):
\begin{equation}
    \dfrac{d}{dx}\csc{f(x)} = \dfrac{d}{dx} \dfrac{1}{\sin(f(x))} = \dfrac{d}{dx} h(g(f(x)))
\end{equation}
where, $h(v) = v^{-1}$, $g(v) = \sin(v)$.
Answer:
\begin{align}
   \dfrac{d}{dx} h(g(f(x))) & = f'(x)\cdot g'(f(x))\cdot h'(g(f(x))) = \\
   & = f'(x)\cdot \cos(f(x))\cdot (-1)\sin^{-2}(f(x))
\end{align}
A: We know that $\displaystyle\frac{d}{dx}\csc(x) = -\csc(x)\cot(x)$
$\displaystyle\frac{d}{dx}\csc(f(x)) = -\csc(f(x))\cot(f(x))$
But this is still incomplete, we still have to use the chain rule and multiply it with the derivative of $f(x)$
Hence
$\displaystyle\frac{d}{dx}\csc(f(x)) = -csc(f(x))\cot(f(x)) f'(x)$
In the first line $f'(x) = 1$ since $f(x) = x$.
A: $$\frac{\text{d}}{\text{dx}}\csc(f(x))=\frac{\text{d}}{\text{dx}}\csc(x)\vert_{f(x)}\cdot \frac{\text{d}}{\text{dx}}f(x)=-\csc(f(x))\cdot\cot(f(x))\cdot f'(x),$$ since $\frac{\text{d}}{\text{dx}}\csc(x)=-\csc(x)\cot(x).$
