Derivative of ${\cot }^{-1}\mathrm{x}$ with respect to ${\tan }^{-1}\left(\frac{1}{\mathrm{x}}\right)$ is

Suppose that $f\left(x\right)=co{t}^{-1}x,g\left(x\right)={\tan }^{-1}\left(\frac{1}{x}\right)$

The derivative of $f\left(x\right)$ with respect to $g\left(x\right)$ is $\begin{array}{rcl} & & \frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}\end{array}$

$\begin{array}{rcl}\frac{f\text{'}\left(x\right)}{g\text{'}\left(x\right)}& =& -\frac{1}{1+x^2}·\frac{1+{\displaystyle \frac{1}{x^2}}}{1}\left(-\frac{1}{x^2}\right)\\ & =& 1\end{array}$