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

The correct option is D $2$

We have, $y=({\tan }^{-1}x{)}^2$

Differentiate using chain rule,

$y_1=\left(2{\tan }^{-1}x\right)\cdot \frac{1}{1+x^2}=\frac{2{\tan }^{-1}x}{1+x^2}$.....(1)

Again differentiate using $\left(u/v\right)$ formula,

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

$\Rightarrow (1+x^2{)}^2{y}_2=2-4x{\tan }^{-1}x$

$\Rightarrow (1+x^2{)}^2{y}_2=2-2x{y}_1(1+x^2)$, using (1)

$\therefore (1+x^2{)}^2{y}_2+2x(x^2+1)y_1=2$