Question

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

Answer 1

Consider the given function

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

Then,

Suppose that

$x^2+\frac{1}{x^2+\frac{1}{x^2+....\infty }}=y$ and we get,

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

$\Rightarrow y^2=x^{2}y+1.........\left(1\right)$

$\Rightarrow x^2=\frac{y^2-1}{y}.......\left(2\right)$

On differentiating this equation with respect to $x$ and we get,

$\Rightarrow 2y\frac{dy}{dx}=x^2\frac{dy}{dx}+y\frac{d}{dx}{x}^2+\frac{d}{dx}1$

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

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

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

Put the value of $x^2$ bye equation (2) and we get,

$\Rightarrow \frac{dy}{dx}=\frac{2xy}{2y-\left(\frac{y^2-1}{y}\right)}$

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

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

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

Hence, this is the answer.