Question

If y = (tan⁻¹ x)², then prove that (1 + x²)² y₂ + 2x(1 + x²)y₁ = 2 ?

Answer 1

Here,

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

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

\[ y_1 = \frac{2 \tan^{- 1} x}{1 + x^2}\]

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

\[ y_2 = \frac{2 - 4x \tan^{- 1} x}{\left( 1 + x^2 \right)^2}\]

\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} - \frac{2 \tan^{- 1} x \times 2x}{\left( 1 + x^2 \right)^2}\]

\[ \Rightarrow y_2 = \frac{2}{\left( 1 + x^2 \right)^2} - \frac{2x y_1}{\left( 1 + x^2 \right)}\]

\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 = 2 - 2x\left( 1 + x^2 \right) y_1 \]

\[ \Rightarrow \left( 1 + x^2 \right)^2 y_2 + 2x\left( 1 + x^2 \right) y_1 = 2\]

Hence proved.