# Solve this

Question:

If $\sqrt{y+x}+\sqrt{y-x}=c$, show that $\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$

Solution:

We are given with an equation $\sqrt{y+x}+\sqrt{y-x}=c$, we have to prove that $\frac{d y}{d x}=\frac{y}{x}-\sqrt{\frac{y^{2}}{x^{2}}-1}$ by using the

given equation we will first find the value of $\frac{d y}{d x}$ and we will put this in the equation we have to prove, so by differentiating the equation on both sides with respect to $x$, we get,

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

$\frac{\sqrt{y-x}+\sqrt{y-x} \frac{d y}{d x}+\sqrt{y+x} \frac{d y}{d x}-\sqrt{y+x}}{2 \sqrt{y+x} \sqrt{y-x}}=0$

$\sqrt{y-x}+\sqrt{y-x} \frac{d y}{d x}+\sqrt{y+x} \frac{d y}{d x}-\sqrt{y+x}=0$

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

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

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

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

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

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

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

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

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