Question

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

• A. $\frac{1}{y}\left(\frac{d y}{d x}\right)^{2}$
• B. $\frac{-c^{4}}{4 y^{3}}$
• C. $y\left(\frac{d y}{d x}\right)^{2}$
• D. $\frac{-c^{2}}{4 y^{3}}$

Answer 1

Answer: (B)

It is given that
$\sqrt{x+y}-\sqrt{x-y}=c$
On squaring both sides, we get
$(x+y)+(x-y)-2 \sqrt{x^{2}-y^{2}} =c^{2} $
$\Rightarrow 2 x-c^{2} =2 \sqrt{x^{2}-y^{2}}$
Again on squaring both sides, we get
$4 x^{2}+c^{4}-4 x c^{2} =4 x^{2}-4 y^{2} $
$\Rightarrow c^{4}-4 x c^{2} =-4 y^{2} $
$\Rightarrow 4 y^{2} =4 x c^{2}-c^{4}$
On differentiating w.r.t. $x$, we get
$8 y \frac{d y}{d x}=4 c^{2}$
$\Rightarrow 2 y \frac{d y}{d x}=c^{2}\,\,\,...(i)$
On differentiating both sides w.r.t. $x$, we get
$2\left(\frac{d y}{d x}\right)^{2}+2 y \frac{d^{2} y}{d x^{2}}=0$
$\Rightarrow \frac{d^{2} y}{d x^{2}}=-\frac{\left(\frac{d y}{d x}\right)^{2}}{y}=-\frac{\left(\frac{c^{2}}{2 y}\right)^{2}}{y} \,\,\,[$ From Eq.(i)]
$=-\frac{c^{4}}{4 y^{3}}$