Find the derivative of tan^{-1}x with respect to log x; (where x ∈ (1, ∞)).

Question

Find the derivative of \(\tan^{-1} x\) with respect to log x; (where \(x \in (1, \infty)\)).

Answer 1

\({y}=\tan ^{-1} {x} \) and \({z}=\log _{{e}} {x}\)

Then \(\frac{d y}{d x}=\frac{{1}}{1+{x}^{2}}\)

and \(\frac{{dz}}{{d x}}=\frac{1}{{x}}\)

So,

\(\frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\)

\(=\frac{\frac{1}{1+\mathrm{x}^{2}}}{\frac{1}{\mathrm{x}}}\)

\(=\frac{\mathrm{x}}{1+\mathrm{x}^{2}}\)