Question

$y=x^2+\frac{1}{x^2+\frac{1}{\frac{1}{x^2+\cdots }}}$, then $\frac{dy}{dx}$

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

Answer 1

Answer: (B)

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

The given equation can be written in the following way as it has infinite terms:

$y=x^2+\frac{1}{y}$

$\Rightarrow y^2=x^{2}y+1$

Differentiating once w.r.t. to x we get,

$\Rightarrow 2y\frac{dy}{dx}=x^2\frac{dy}{dx}+2xy$

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

$\Rightarrow \frac{dy}{dx}=\frac{2xy}{2y-x^2}$ ....Answer