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

The Correct Option is 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
\(\left(1+x^{2}\right)^{2}\left(\frac{d^{2} y}{d x^{2}}\right)+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2\)

So, the correct option is (B): 2