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
$\frac{dy}{dx}=2\left({\tan }^{-1}x\right)\times \frac{d}{dx}\left({\tan }^{-1}x\right)$
$\Rightarrow \frac{dy}{dx}=\frac{2{\tan }^{-1}x}{1+x^2}$
$\left(1+x^2\right)\frac{dy}{dx}=2{\tan }^{-1}x$
Again, differentiating both sides w.r.t. $x$, we get
$\left(1+x^2\right)\frac{d}{dx}\left(\frac{dy}{dx}\right)+\frac{dy}{dx}\frac{d}{dx}\left(1+x^2\right)$
$=2\frac{d}{dx}\left({\tan }^{-1}x\right)$
$\Rightarrow \left(1+x^2\right)\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}(0+2x)=\frac{2}{1+x^2}$
$\Rightarrow {\left(1+x^2\right)}^2\frac{d^{2}Y}{d{x}^2}+2x\left(1+x^2\right)\frac{dy}{dx}=2$
or ${\left(1+x^2\right)}^2{y}_2+2x\left(1+x^2\right)y_1=2$