Question

please solve this please ​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow \frac{dy}{dx}=\frac{2xy^2}{y^2+1}$

Explanation:

Given :

$\displaystyle \sf y=x^2+ \cfrac{1}{ x^2+ \cfrac{1}{x^2+ \cfrac{1}{ x^2+ \cfrac{1}{x^2+.... \infty}}}}}$

Clearly we can see all vales are same as in denominator if we let y :

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

$\displaystyle \sf \longrightarrow y-\frac{1}{y}=x^2$

$\displaystyle \sf \longrightarrow y-y^{-1}=x^2$

Now diff. w.r.t. x :

$\displaystyle \sf \longrightarrow \frac{dy}{dx} -\left(-\frac{1}{y^2}.\frac{dy}{dx}\right)=\frac{d}{dx} (x^2)$

$\displaystyle \sf \longrightarrow \frac{dy}{dx} +\left(\frac{1}{y^2}.\frac{dy}{dx}\right)=2x$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}\left(1+\frac{1}{y^2}\right)=2x$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}\left(\frac{y^2+1}{y^2}\right)=2x$

$\displaystyle \sf \longrightarrow \frac{dy}{dx}=\frac{2xy^2}{y^2+1}$

Hence we get required answer.