Question

If $y = (\tan^{-1}x)^2$ then $(x^2+1)^2 y_2+2x(x^2+1)y_1$ is equal to

• A. 0
• B. 1
• C. 4
• D. 2

Answer 1

Answer: (D)

Given, $y=\left(\tan ^{-1} x\right)^{2}$
On differentiating both sides w.r.t. $x$, we get
$ \frac{d y}{d x}=2\left(\tan ^{-1} x\right) \times \frac{d}{d x}\left(\tan ^{-1} x\right)$
$\Rightarrow \frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}$
$\left(1+x^{2}\right) \frac{d y}{d x}=2 \tan ^{-1} x$
Again, differentiating both sides w.r.t. $x$, we get
$\left(1+x^{2}\right) \frac{d}{d x}\left(\frac{d y}{d x}\right)+\frac{d y}{d x} \frac{d}{d x}\left(1+x^{2}\right)$
$=2 \frac{d}{d x}\left(\tan ^{-1} x\right)$
$\Rightarrow \left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}(0+2 x)=\frac{2}{1+x^{2}}$
$\Rightarrow \left(1+x^{2}\right)^{2} \frac{d^{2} Y}{d x^{2}}+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2$
or $\left(1+x^{2}\right)^{2} y_{2}+2 x\left(1+x^{2}\right) y_{1}=2$