# Derivative of Cotangent Function

From ProofWiki

## Theorem

- $\displaystyle D_x \left({\cot x}\right) = -\csc^2 x = \frac {-1} {\sin^2 x}$, when $\sin x \ne 0$.

## Proof

- From the definition of the cotangent function, $\cot x = \dfrac {\cos x} {\sin x}$.
- From Derivative of Sine Function we have $D_x \left({\sin x}\right) = \cos x$.
- From Derivative of Cosine Function we have $D_x \left({\cos x}\right) = -\sin x$.

Then:

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle D_x \left({\cot x}\right)\) | \(=\) | \(\displaystyle \) | \(\displaystyle \frac {\sin x \left({-\sin x}\right) - \cos x \cos x} {\sin^2 x}\) | \(\displaystyle \) | \(\displaystyle \) | Quotient Rule for Derivatives | ||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \frac {-\left({\sin^2 x + \cos^2 x}\right)} {\sin^2 x}\) | \(\displaystyle \) | \(\displaystyle \) | |||

\(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle \) | \(\displaystyle \frac {-1} {\sin^2 x}\) | \(\displaystyle \) | \(\displaystyle \) | Sum of Squares of Sine and Cosine |

This is valid only when $\sin x \ne 0$.

The result follows from the definition of the cosecant function.

$\blacksquare$