Question

Find the second order derivatives of the function
\(tan^{-1}x\)

Answer 1

The correct answer is \(=\frac{-2x}{(1+x^2)^2}\)
Let \(y=tan^{-1}x\)
Then,
\(\frac{dy}{dx}=\frac{d}{dx}(tan^{-1}x)=\frac{1}{1+x^2}\)
\(∴\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{1}{1+x^2}]\)
\(=\frac{d}{dx}(1+x^2)^{-1}\)
\(=(-1).(1+x^2)^{-2}.\frac{d}{dx}(1+x^2)\)
\(=\frac{-1}{(1+x^2)^2}\times 2x\)
\(=\frac{-2x}{(1+x^2)^2}\)