JEE Main 2017 — Differentiation Question with Solution

The given equation is

$$y=(x^2+\sqrt{x^2-1}{)}^{15}+(x-\sqrt{x^2-1}{)}^{15}$$

Differentiating w.r.t. x, we get

$$\frac{dy}{dx}=15(x+\sqrt{x^2-1}{)}^{14}\left(1+\frac{1(2x)}{2\sqrt{x^2-1}}\right)+15(x-\sqrt{x^2-1}{)}^{14}\left(1-\frac{1(2x)}{2\sqrt{x^2-1}}\right)$$

Here, we have used the standard differentiatials

$$\frac{d}{dx}{x}^n=n{x}^{n-1}$$

That is, $$\frac{d}{dx}(\sqrt{f(x)})=\frac{1}{2\sqrt{f(x)}}\times \frac{d}{dx}(f(x))$$

Therefore

$$\frac{dy}{dx}=\frac{15{(x+\sqrt{x^2-1})}^{14}(\sqrt{x^2-1}+x)}{\sqrt{x^2-1}}+\frac{15{(x-\sqrt{x^2-1})}^{14}(\sqrt{x^2-1}-x)}{\sqrt{x^2-1}}$$

$$\Rightarrow \sqrt{x^2-1}\frac{dy}{dx}=15(x+\sqrt{x^2-1}{)}^{15}-15(x-\sqrt{x^2-1}{)}^{15}$$

Differentiating w.r.t. x, we get

$$\frac{1(2x)}{2\sqrt{x^2+1}}\frac{dy}{dx}+\sqrt{x^2-1}\frac{d^{2}y}{d{x}^2}=15\times 15(x+\sqrt{x^2-1}{)}^{14}\left(1+\frac{1(2x)}{2\sqrt{x^2-1}}\right)-15\times 15(x-\sqrt{x^2-1}{)}^{14}\left(\frac{1-1(2x)}{2\sqrt{x^2-1}}\right)$$

$$\Rightarrow \frac{x}{\sqrt{x^2-1}}\frac{dy}{dx}+\sqrt{x^2-1}\frac{d^{2}y}{d{x}^2}=225(x+\sqrt{x^2-1}{)}^{14}\frac{(\sqrt{x^2-1}+x)}{\sqrt{x^2-1}}-\frac{225{(x-\sqrt{x^2-1})}^{14}(\sqrt{x^2-1}-x)}{\sqrt{x^2-1}}$$

$$\Rightarrow \sqrt{x^2-1}\left[\frac{x}{\sqrt{x^2-1}}\frac{dy}{dx}+\sqrt{x^2-1}\frac{d^{2}y}{d{x}^2}\right]=225(x+\sqrt{x^2-1}{)}^{15}+225(x-\sqrt{x^2-1}{)}^{15}$$

$$\Rightarrow x\frac{dy}{dx}+(x^2-1)\frac{d^{2}y}{d{x}^2}=225\left[{(x+\sqrt{x^2-1})}^{15}+{(x-\sqrt{x^2-1})}^{15}\right]$$

Substituting $$(x+\sqrt{x^2-1}{)}^{15}+(x-\sqrt{x^2-1}{)}^{15}=y$$, we get

$$(x^2-1)\frac{d^{2}y}{d{x}^2}+\frac{xdy}{dx}=225y$$