Question

If $y={\left({\tan }^{-1}x\right)}^2$ then ${\left(x^2+1\right)}^2\frac{d^{2}y}{d{x}^2}+2x\left(x^2+1\right)\frac{dy}{dx}=$

• A. 4
• B. 2
• C. 1
• D. 0

Answer 1

Answer: (B)

We have,
$y={\left({\tan }^{-1}x\right)}^2$
On differentiating w.r.t. $x,$ we get
$\frac{dy}{dx}=\frac{2{\tan }^{-1}x}{1+x^2}$
$\Rightarrow \left(1+x^2\right)\frac{dy}{dx}=2{\tan }^{-1}x$
On squaring both sides, we get
${\left(1+x^2\right)}^2{\left(\frac{dy}{dx}\right)}^2=4{\left({\tan }^{-1}x\right)}^2$
$\Rightarrow {\left(1+x^2\right)}^2{\left(\frac{dy}{dx}\right)}^2=4y\left[\because y={\left({\tan }^{-1}x\right)}^2\right]$
Again, differentiating w.r.t.x, we get
${\left(1+x^2\right)}^2\left(2\frac{dy}{dx}\cdot \frac{d^{2}y}{d{x}^2}\right)+2\left(1+x^2\right)(2x){\left(\frac{dy}{dx}\right)}^2=4\frac{dy}{dx}$
On dividing both sides by $2\frac{dy}{dx}$,
we get
${\left(1+x^2\right)}^2\left(\frac{d^{2}y}{d{x}^2}\right)+2x\left(1+x^2\right)\frac{dy}{dx}=4$