Question

Let $y=f\left(x\right)$ is a positive function which satisfies equation $\sqrt{y^2+2x}+\sqrt{y^2-2x}=2x^2,\text{then}\frac{dy}{dx}$ is

• A. $\frac{2x^3-x^{-3}}{y}$
• B. $\frac{2x^4-x^{-2}}{y}$
• C. $\frac{2x^4-x^{-2}}{\sqrt{x^6+1}}$
• D. $\frac{2x^3-x^{-3}}{\sqrt{1+x^6}}$

Answer 1

Answer: (A), (C)

$\frac{2x^4-x^{-2}}{\sqrt{x^6+1}}$

$\sqrt{y^2+2x}+\sqrt{y^2-2x}=2x^2$...(1)
rationalizing the equation (1) we get
$\sqrt{y^2+2x}-\sqrt{y^2-2x}=\frac{2}{x}$ ...(2)
adding the equations (1) & (2)
$2\sqrt{y^2+2x}=2x^2+\frac{2}{x}$
$(y^2+2x)=(x^2+\frac{1}{x}{)}^2$
$y^2=x^4+\frac{1}{x^2}\Rightarrow y=\sqrt{x^4+\frac{1}{x^2}}$
$2y.\frac{dy}{dx}=4x^3-2x^{-3}$
$\frac{dy}{dx}=\frac{2x^3-x^{-3}}{y}$
$\frac{dy}{dx}=\frac{2x^3-x^{-3}}{\sqrt{\frac{1+x^6}{x^2}}}=\frac{2x^4-x^{-2}}{\sqrt{1+x^6}}$