Question

Differenciate log x with respect to cot x

Answer 1

if you know the chain rule


`\displaystyle(df)/(dx) = (df)/(dg) (dg)/(dx) \implies (df)/(dg) = ( (df)/(dx) )/( (dg)/(dx) )`

hence


`\displaystyle (d \log x)/(d \cot x) = ( (d \log x)/(dx) )/( (d \cot x)/(dx) ) = ((1)/(x \ln 10))/(-\csc^2 x) = - (1)/(x \ln 10 \cdot \csc^2 x)`

Assuming that
`\log x = \log_(10) x`