Question

If $y=x^2+\frac{1}{x^2+{\displaystyle \frac{1}{x^2+{\displaystyle \frac{1}{x^2.....\infty }}}}}$ 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. $\frac{2x}{2+{\displaystyle \frac{x^2}{y}}}$

Answer 1

The correct option is A

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

Explanation for the correct option.

Finding the value of $\frac{dy}{dx}$:

In $y=x^2+\frac{1}{x^2+{\displaystyle \frac{1}{x^2+{\displaystyle \frac{1}{x^2.....\infty }}}}}$, $x^2+\frac{1}{x^2+{\displaystyle \frac{1}{x^2.....\infty }}}$ is being repeated and is equal to $y$.

So, we can say that

$\begin{array}{rcl}y& =& x^2+\frac{1}{y}\\ & \Rightarrow & y^2=1+x^{2}y\end{array}$

By differentiating with respect to $x$, we get

$\begin{array}{rcl}2y\frac{dy}{dx}& =& 2xy+x^2\frac{dy}{dx}\\ \Rightarrow \frac{dy}{dx}& =& \frac{2xy}{2y-x^2}\end{array}$

Hence, option A is correct.