 # What is the derivative of f(x)=\cot(x)? Lorraine Harvey 2021-12-18 Answered
What is the derivative of $f\left(x\right)=\mathrm{cot}\left(x\right)$?
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$\mathrm{cot}\left(x\right)=\frac{1}{\mathrm{tan}\left(x\right)}$, so ${f}^{\prime }\left(x\right)=\frac{1}{\mathrm{tan}}dx$
The quotion rule:
$d\left(\frac{g\left(x\right)}{h\left(x\right)}\right)=\left(\frac{{g}^{\prime }\left(x\right)h\left(x\right)-g\left(x\right){h}^{\prime }\left(x\right)}{{g\left(x\right)}^{2}}\right)dx$,
So, we have,
$g\left(x\right)=1$, $h\left(x\right)=\mathrm{tan}\left(x\right)$, ${g}^{\prime }\left(x\right)=0$, ${h}^{\prime }\left(x\right)={\mathrm{sec}}^{2}\left(x\right)$
Thus,
$\frac{0×\mathrm{tan}\left(x\right)-1×{\mathrm{sec}}^{2}\left(x\right)}{{\mathrm{tan}}^{2}\left(x\right)}=-\frac{{\mathrm{sec}}^{2}\left(x\right)}{{\mathrm{tan}}^{2}\left(x\right)}=-{\mathrm{csc}}^{2}\left(x\right)$
###### Not exactly what you’re looking for? Natalie Yamamoto
Derivative of $\mathrm{cot}\left(x\right)$ is equal to $-{\mathrm{csc}}^{2}\left(x\right)$.