Question

Differenciate log x with respect to cot x

Answer 1

If you know the chain rule

$\displaystyle\frac{df}{dx} = \frac{df}{dg} \frac{dg}{dx} \implies \frac{df}{dg} = \frac{ \frac{df}{dx} }{ \frac{dg}{dx} }$

hence

$\displaystyle \frac{d \log x}{d \cot x} = \frac{ \frac{d \log x}{dx} }{ \frac{d \cot x}{dx} } = \frac{\frac{1}{x \ln 10}}{-\csc^2 x} = - \frac{1}{x \ln 10 \cdot \csc^2 x}$

Assuming that $\log x = \log_{10} x$