Derivative of arccot

Theorem

The following formula holds: $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{z^2+1},$$ where $\mathrm{arccot}$ denotes the inverse cotangent function.

Proof

If $y=\mathrm{arccot}(z)$ then $\cot(y)=z$. Now use implicit differentiation with respect to $z$ to get $$-\csc^2(y)y'=1.$$ Substituting back in $y=\mathrm{arccos}(z)$ yields the formula $$\dfrac{\mathrm{d}}{\mathrm{d}z} \mathrm{arccot}(z) = -\dfrac{1}{\csc^2(\mathrm{arccot}(z))} = -\dfrac{1}{z^2+1},$$ as was to be shown. █