Question

if $y=x^2+\left(\frac{1}{x^2+\frac{1}{x^2+....\infty }}\right)$, show that $\frac{dy}{dx}=\frac{2x{y}^2}{y^2+1}$.

Answer 1

We have,

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

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

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

$\Rightarrow y-x^2=\frac{1}{y}$

$\Rightarrow y^1-2x=\frac{-1}{y^2}{y}^1$

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

Hence, proved.