#### need solution for RD Sharma maths class 12 chapter Differentiation exercise 10.4 question 31

$\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$     At $x=\frac{\pi}{4}$

Hint:

Use chain rule

Given:

$\sqrt{y+x}+\sqrt{y-x}=c$

Solution:

$\sqrt{y+x}+\sqrt{y-x}=c$

Differentiate the given equation w.r.t x

$\frac{d}{d x}(\sqrt{y+x}+\sqrt{y-x})=\frac{d(c)}{d x}$

$\frac{d(\sqrt{y+x})}{d x}+\frac{d(\sqrt{y-x})}{d x}=0$                        $\left[\because \frac{d(\operatorname{cons} \tan t)}{d x}=0\right]$

$\frac{d(\sqrt{y+x})}{d(y+x)} \times \frac{d(y+x)}{d x}+\frac{d(\sqrt{y-x})}{d(y-x)} \times \frac{d(y-x)}{d x}=0$            [Using chain rule]

$\frac{1}{2 \sqrt{y+x}} \times\left(\frac{d y}{d x}+\frac{d x}{d x}\right)+\frac{1}{2 \sqrt{y-x}} \times\left(\frac{d y}{d x}-\frac{d x}{d x}\right)=0$            $\left[\because \frac{d\left(x^{n}\right)}{d x}=n x^{n-1}\right]$

$\frac{1}{2 \sqrt{y+x}}\left(\frac{d y}{d x}+1\right)+\frac{1}{2 \sqrt{y-x}}\left(\frac{d y}{d x}-1\right)=0$

$\frac{d y}{d x}\left(\frac{1}{2 \sqrt{y+x}}+\frac{1}{2 \sqrt{y-x}}\right)=\frac{1}{2 \sqrt{y-x}}-\frac{1}{2 \sqrt{y+x}}$

$\frac{d y}{d x}\left(\frac{\sqrt{y-x}+\sqrt{y+x}}{2 \sqrt{y+x} \cdot \sqrt{y-x}}\right)=\frac{\sqrt{y+x}-\sqrt{y-x}}{(2 \sqrt{y-x} \cdot \sqrt{y+x})}$

$\frac{d y}{d x} \times[\sqrt{y-x}+\sqrt{y+x}]=[\sqrt{y+x}-\sqrt{y-x}]$

$\frac{d y}{d x}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}$

Rationalizing the denominator

$\frac{d y}{d x}=\frac{(\sqrt{y+x}-\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}{(\sqrt{y+x}+\sqrt{y-x})(\sqrt{y+x}-\sqrt{y-x})}$

$=\frac{(\sqrt{y+x}-\sqrt{y-x})^{2}}{(\sqrt{y+x})^{2}-(\sqrt{y-x})^{2}}$                $\left[\because a^{2}-b^{2}=(a+b)(a-b)\right]$

$\frac{d y}{d x}=\frac{(\sqrt{y+x})^{2}+(\sqrt{y-x})^{2}-2 \sqrt{y+x} \cdot \sqrt{y-x}}{(y+x)-(y-x)}$

$=\frac{y+x+y-x-2 \sqrt{y+x} \cdot \sqrt{y-x}}{y+x-y+x}$

$=\frac{2 y-2 \sqrt{y+x} \cdot \sqrt{y-x}}{2 x}$

$=\frac{2(y-\sqrt{y+x} \cdot \sqrt{y-x})}{2 x}$

$=\frac{y-\sqrt{y+x} \cdot \sqrt{y-x}}{x}$

$=\frac{y-\sqrt{y^{2}-x^{2}}}{x}$

$=\frac{y}{x}-\frac{\sqrt{y^{2}-x^{2}}}{\sqrt{x^{2}}}$

$\therefore \frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$

Thus proved