Question

If $y=\sqrt{x+\sqrt{y+\sqrt{x+......}}}$, then $\frac{dy}{dx}=$

• A. $\frac{x-y^2}{2y^3-2xy+1}$
• B. $\frac{-x+y^2}{2y^3-2xy-1}$
• C. $\frac{x+y^2}{2y^3-2xy-1}$
• D. None of these

Answer 1

Answer: (B)

$\frac{-x+y^2}{2y^3-2xy-1}$

$y=\sqrt{x+\sqrt{y+\sqrt{x+.......}}}$

$y=\sqrt{x+\sqrt{y+y}}[\because y=\sqrt{x+\sqrt{y+\sqrt{x+.......}}}]$

Squaring both sides, we have

$y^2=x+\sqrt{2y}$

$\Rightarrow y^2-x=\sqrt{2y}$

Again squaring both sides, we have

$y^4+x^2-2y^{2}x=2y$

$y^4+x^2-2y-2y^{2}x=0$

Differentiating above equation w.r.t. $x$, we have

$4y^3\frac{dy}{dx}+2x-2\frac{dy}{dx}-2(2xy\frac{dy}{dx}+y^2)=0$

$4y^3\frac{dy}{dx}+2x-2\frac{dy}{dx}-4xy\frac{dy}{dx}-2y^2=0$

$\frac{dy}{dx}(4y^3-4xy-2)=2y^2-2x$

$\frac{dy}{dx}=\frac{(y^2-x)}{(2y^3-2xy-1)}$

Hence the correct answer is $\frac{dy}{dx}=\frac{(y^2-x)}{(2y^3-2xy-1)}$.