Question

If $y=\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+ln\sqrt{x+\sqrt{x^2+1}}$ find the value of $x\frac{dy}{dx}+\ln \frac{dy}{dx}$

• A. $2y$
• B. $3y$
• C. $y$
• D. $4y$

Answer 1

The correct option is A $2y$

Given,

$y=\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+\ln \left(\sqrt{x+\sqrt{x^2+1}}\right)$

$\frac{dy}{dx}=\frac{d}{dx}(\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+\ln \left(\sqrt{x+\sqrt{x^2+1}}\right))$

$=\frac{d}{dx}\left(\frac{x^2}{2}\right)+\frac{d}{dx}\left(\frac{1}{2}x\sqrt{x^2+1}\right)+\frac{d}{dx}\left(\ln \left(\sqrt{x+\sqrt{x^2+1}}\right)\right)$

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

$=x+\sqrt{x^2+1}$

Now,

$x\frac{dy}{dx}+\ln \frac{dy}{dx}$

$=x(x+\sqrt{x^2+1})+\ln (x+\sqrt{x^2+1})$

$=x^2+x\sqrt{x^2+1}+\ln (x+\sqrt{x^2+1})$

$=2(\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+ln\sqrt{x+\sqrt{x^2+1}})$

$=2y$