Discussion Forum

1)

If y= $(tan^{-1} x)^{2}$ , then

 $(x^{2} +1)^{2}\frac{d^{2}y}{dx^{2}}+2x(x^{2}+1)\frac{dy}{dx}=$

Answer:

Option A

Explanation:

We have   y= $(tan^{-1} x)^{2}$

 on differentiating  w.r.t x, we get

$\frac{dy}{dx}=\frac{2 \tan^{-1}x}{1+x^{2}}$

 $\Rightarrow$    $(1+x^{2})\frac{dy}{dx}=2 \tan^{-1}x$

 On squaring both sides, we get

$(1+x^{2})^{2}(\frac{dy}{dx})^{2}=4( \tan^{-1}x)^{2}$

 $\Rightarrow$    $(1+x^{2})^{2}(\frac{dy}{dx})^{2}=4y$       $[\because y=  \tan^{-1}x)^{2}]$

 Again , differentiating w.r.t x , we get

$(1+x^{2})^{2}\left(2\frac{dy}{dx}.\frac{d^2y}{d^2x}\right)+2(1+x^{2})(2x)\left(\frac{dy}{dx}\right)^{2}=4\frac{dy}{dx} $

 On dividing both sides by $2\frac{dy}{dx}$,

we get

$(1+x^{2})^{2}\left(\frac{d^2y}{d^2x}\right)+2x(1+x^{2})\frac{dy}{dx}=4$