Question

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

Answer 1

We have,

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

On differentiating w.r.t $x$, we get

$\frac{1}{2\sqrt{y+x}}(1+\frac{dy}{dx})+\frac{1}{2\sqrt{y-x}}(\frac{dy}{dx}-1)=0$

$\frac{dy}{dx}[\frac{1}{\sqrt{y+x}}+\frac{1}{\sqrt{y-x}}]=\frac{1}{\sqrt{y-x}}-\frac{1}{\sqrt{y+x}}$

$\frac{dy}{dx}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}=\frac{1-\sqrt{\frac{y-x}{y+x}}}{1+\sqrt{\frac{y-x}{y+x}}}$

$\frac{dy}{dx}=\frac{1-\sqrt{\frac{y-x}{y+x}}}{1+\sqrt{\frac{y-x}{y+x}}}=\frac{\sqrt{y+x}-\sqrt{y-x}}{\sqrt{y+x}+\sqrt{y-x}}$

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

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

Hence, proved.