Derivative of cotangent

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}x}\cot(x)=-\csc^2(x),$$ where $\cot$ denotes the cotangent and $\csc$ denotes the cosecant.

Proof

Apply the quotient rule to the definition of cotangent using derivative of sine and derivative of cosine to see $$\begin{array}{ll} \dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) &= \dfrac{\mathrm{d}}{\mathrm{d}x} \left[ \dfrac{\cos(x)}{\sin(x)} \right] \\ &= \dfrac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)} \\ &= -\dfrac{\sin^2(x)+\cos^2(x)}{\sin^2(x)}. \end{array}$$ Now apply the Pythagorean identity for sin and cos and the definition of cosecant to see $$\dfrac{\mathrm{d}}{\mathrm{d}x} \cot(x) = -\dfrac{1}{\sin^2(x)} = -\csc^2(x),$$ as was to be shown. █