Question

If $\sqrt{y+x}+\sqrt{y-x}=c$, then $\frac{dy}{dx}$ is equal to

• A. $\frac{2x}{c^2}$
• B. $\frac{x}{y+\sqrt{y^2-x^2}}$
• C. $\frac{y-\sqrt{y^2-x^2}}{x}$
• D. $\frac{c^2}{2y}$

Answer 1

Answer: (B), (C)

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

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

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

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

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

(By rationalizing Numerator or Denominator)

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

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