Question

If y = tan⁻¹ x, show that \[\left( 1 + x^2 \right) \frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\] ?

Answer 1

Here,

\[y = \tan^{- 1} x\]

\[\text { Differentiating w . r . t . x, we get }\]

\[\frac{d y}{d x} = \frac{1}{1 + x^2}\]

\[\text { Differentiating again w . r . t . x, we get }\]

\[\frac{d^2 y}{d x^2} = \frac{- 2x}{\left( 1 + x^2 \right)^2}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- 2x}{1 + x^2} \times \frac{1}{1 + x^2}\]

\[ \Rightarrow \frac{d^2 y}{d x^2} = \frac{- 2x\frac{dy}{dx}}{1 + x^2}\]

\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} = - 2x\frac{dy}{dx}\]

\[ \Rightarrow \left( 1 + x^2 \right)\frac{d^2 y}{d x^2} + 2x\frac{dy}{dx} = 0\]

Hence proved.