The derivative of $\csc x$ is $\boxed{-\csc x\cot x}$. We will use our knowledge of the derivatives of $\sin x$ and $\cos x$ to prove this result. Recall that

\begin{align*} \dfrac d{dx} \sin x = \cos x\end{align*}

and

\begin{align*} \dfrac d{dx} \cos x = -\sin x\end{align*}

For a detailed discussion and proof of the derivative of $\cos x$, click here. For a discussion and proof proof of the derivative of $\sin x$, click here. Since

\begin{align*} \csc x = \frac{1}{\sin x}\end{align*}

it makes sense to use the quotient rule, which states that for two functions $f(x)$ and $g(x)$

\begin{align*} \frac d{dx}\bigg(\frac{f(x)}{g(x)}\bigg) = \frac{f'(x) g(x) – g'(x) f(x)}{g^2(x)}\end{align*}

In this case, we will let $f(x) = 1$ and $g(x) = \sin x$. Then $f'(x) = 0$ and $g'(x) = \cos x$. We will now substitute these values into the quotient rule to get

\begin{align*} \frac d{dx}(\csc x) = \frac d{dx}\bigg(\frac{1}{\sin x}\bigg) = \frac{0\cdot\sin x – (\cos x)1}{\sin^2 x} = \frac{-\cos x}{\sin^2 x}\end{align*}

At this point, we recall that

\begin{align*} \frac{\cos x}{\sin x} = \cot x\end{align*}

and

\begin{align*} \frac{1}{\sin x} = \csc x\end{align*}

In conclusion, we combine these two facts to get

\begin{align*} \frac d{dx}(\csc x) = -\frac{\cos x}{\sin^2 x} = \frac{1}{\sin x}\cdot\frac{\cos x}{\sin x} = -\csc x\cot x\end{align*}

Similarly, we may use this technique to find the derivative of other trigonometric functions, like $\tan x$, $\sec x$,  and $\cot x$. You can find the detailed steps for finding the derivative of $\tan x$ here. Similarly, the steps for finding the derivative of $\cot x$ are here and for  $\sec x$ are here.