# Derivative of $\csc(f(x))$

Could someone explain how can I find the derivative:

$$\frac{d}{dx}\csc[f(x)] = ?$$

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All answers inherently assume f to be differentiable. Pls consider editing it in the body. – Gautam Shenoy Nov 22 '12 at 11:15

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 $f(x) = x$.

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Thanks for the answer. However, I meant the f(x) to be any differentiable function. – General Stubbs Nov 22 '12 at 23:49
Yes for any $f(x)$ differentiable, $\displaystyle\frac{d}{dx}\csc(f(x)) = -\csc(f(x))\cot(f(x))f'(x)$. In the last line what I meant was if you take $f(x) = x$ we get the normal derivative of $csc(x)$ since $f'(x) =1$ :) – Keith Reyes Nov 23 '12 at 2:11

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):

$$\dfrac{d}{dx}\csc{f(x)} = \dfrac{d}{dx} \dfrac{1}{\sin(f(x))} = \dfrac{d}{dx} h(g(f(x)))$$ 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}

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$$\frac{\text{d}}{\text{dx}}\csc(f(x))=\frac{\text{d}}{\text{dx}}\csc(x)\vert_{f(x)}\cdot \frac{\text{d}}{\text{dx}}f(x)=-\csc(f(x))\cdot\cot(f(x))\cdot f'(x),$$ since $\frac{\text{d}}{\text{dx}}\csc(x)=-\csc(x)\cot(x).$

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