Question

If $y=\frac{x^2}{2}+\frac{1}{2}x\sqrt{x^2+1}+\ln \sqrt{x+\sqrt{x^2+1}}$, then the value of $x{y}^{\prime }+\log y^{\prime }$ is

• A. $y$
• B. $2y$
• C. $0$
• D. $-2y$

Answer 1

The correct option is B $2y$

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

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

Differentiating above equation with respect to $x$, we get

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

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

$y^{{}^{\prime }}=\frac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}}\times (\sqrt{x^2+1}+x)+\frac{1}{2\sqrt{x^2+1}}$

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

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

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

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

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

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

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

Therefore,

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

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

Multiplying and dividing the R.H.S by 2, we get

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

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

$x{y}^{{}^{\prime }}+ln{y}^{{}^{\prime }}=2y$

So, the answer is option B