Question

If $\sqrt{y+x}+\sqrt{y-x}=c$ (where $c\ne 0$), then $\frac{dy}{dx}$ has the value 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: (A), (B), (C)

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

$\sqrt{y+x}+\sqrt{y-x}=c$
Squaring both sides, we get
${(\sqrt{y+x}+\sqrt{y-x})}^2=c^2$
$2y+2\sqrt{y^2-x^2}=c^2$
$y+\sqrt{y^2-x^2}=\frac{c^2}{2}$...........(1)
Differentiating bothe the sides, we get
$\frac{dy}{dx}+\frac{1}{2\sqrt{y^2-x^2}}\left(2y\frac{dy}{dx}-2x\right)=0$
$\frac{dy}{dx}+\frac{y}{\sqrt{y^2-x^2}}\frac{dy}{dx}-\frac{x}{\sqrt{y^2-x^2}}=0$
$\frac{dy}{dx}(1+\frac{y}{\sqrt{y^2-x^2}})=\frac{x}{\sqrt{y^2-x^2}}$
$\frac{dy}{dx}=\frac{x}{y+\sqrt{y^2-x^2}}$..................Option (b)
On rationalizing we get,
$\frac{dy}{dx}=\frac{x(y-\sqrt{y^2-x^2})}{x^2}$
$=\frac{y-\sqrt{y^2-x^2}}{x}$........Option (c)
Putting $y+\sqrt{y^2-x^2}=c^2$. we get
$\frac{dy}{dx}=\frac{x}{\frac{c^2}{2}}=\frac{2x}{c^2}$......... Option (a)