Question

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

• A. $\frac{y-x+1}{y-x-1}$
• B. $\frac{y-x}{y+x}$
• C. $\frac{x+y+1}{y-x}$
• D. $\frac{y-x+1}{y+x}$

Answer 1

Answer: (A)

$\frac{y-x+1}{y-x-1}$

$\sqrt{y-\sqrt{y-\sqrt{y...-\infty }}}=\sqrt{x+\sqrt{x+\sqrt{x...+\infty }}}----\left[1\right]$

Squaring both Sides,

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

From $\left[1\right]$

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

$y=x+2\sqrt{x+\sqrt{x+\sqrt{x...+\infty }}}$

$y-x=2\sqrt{x+\sqrt{x+\sqrt{x...+\infty }}}----\left[2\right]$

Squaring both Sides,

${(y-x)}^2=4(x+\sqrt{x+\sqrt{x+\sqrt{x...+\infty }}})$

From $\left[2\right]$

$y^2+x^2-2yx=4(x+\frac{(y-x)}{2})$

$y^2+x^2-2yx=4(\frac{x}{2}+\frac{y}{2})$

Multiplying 2,

$2y^2+2x^2-4yx=4(x+y)$

Differentiating with respect to x,

$4y\frac{dy}{dx}+4x-4y-4x\frac{dy}{dx}=4+4\frac{dy}{dx}$

$\frac{dy}{dx}(4y-4x-4)=4y-4x+4$

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

$\frac{dy}{dx}=\frac{y-x+1}{y-x-1}$