The derivative of $\cot x$ is $\boxed{-\csc^2 x}$. 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*} \cot x = \frac{\cos x}{\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 $f(x) = \cos x$ and $g(x) = \sin x$. Then $f'(x) = -\sin x$ and $g'(x) = \cos x$. We will now substitute these values into the quotient rule to get

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

At this point, we will recall the Pythagorean Identity:

\begin{align*} \cos^2 x + \sin^2 x= 1\end{align*}

Applying this identity yields

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

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