Question

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