Question

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

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

Answer 1

Answer: (A)

$y=x^2+\frac{1}{y}$
or $y^2=x^{2}y+1$
or $2y\frac{dy}{dx}=y\cdot 2x+x^2\frac{dy}{dx}$
or $\frac{dy}{dx}=\frac{2xy}{2y-x^2}$