Question

If $\sqrt{\left(x+y\right)}+\sqrt{\left(y-x\right)}=a$, then $\frac{dy}{dx}=$

• A. $\frac{\sqrt{\left(x+y\right)}-\sqrt{y-x}}{\sqrt{y-x}+\sqrt{x+y}}$
• B. $\frac{2\sqrt{x-y}}{\sqrt{x+y}-\sqrt{x-y}}$
• C. $\frac{x+y+\sqrt{xy}}{\sqrt{x+y}}$
• D. $\frac{x^2+y^2+2xy}{x^2+y^2}$

Answer 1

Answer: (A)

We have, $\sqrt{x+y}+\sqrt{y-x}=a$
Differentiating $w$.$r$.$t$. $x$, we get
$\frac{1}{2\sqrt{x+y}}\left(1+\frac{dy}{dx}\right)+\frac{1}{2\sqrt{y-x}}\left(\frac{dy}{dx}-1\right)=0$
$\Rightarrow \frac{1}{\sqrt{x+y}}-\frac{1}{\sqrt{y-x}}=-\left(\frac{1}{\sqrt{x+y}}+\frac{1}{\sqrt{y-x}}\right)\frac{dy}{dx}$
$\Rightarrow \frac{-\left(\sqrt{y-x}-\sqrt{x+y}\right)}{\left(\sqrt{y-x}+\sqrt{x+y}\right)}=\frac{dy}{dx}$
$\Rightarrow \frac{dy}{dx}=\frac{\sqrt{x+y}-\sqrt{y-x}}{\sqrt{y-x}+\sqrt{x+y}}$