Question

If $y=x^2+\frac{1}{x^2+\frac{1}{x^2+\frac{1}{x^2+...\infty }}}$, then $\frac{dy}{dx}$ is equal to

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

Answer 1

Answer: (A)

$\frac{2xy}{2y-x^2}$

Given equation can be written as
$y=x^2+\frac{1}{y}\Rightarrow y^2=x^{2}y+1$
On differentiating with respect to $x$, we get
$2y\frac{dy}{dx}=x^2\frac{dy}{dx}+2xy+0$
$\Rightarrow \frac{dy}{dx}(2y-x^2)=2xy$
$\Rightarrow \frac{dy}{dx}=\frac{2xy}{2y-x^2}$