Answer
$\displaystyle \frac{dy}{dx}=\frac{2x^{2}+1}{(x^{2}+1)^{1/2}}$
Work Step by Step
$y=x\sqrt{x^{2}+1}$
$y=x\cdot(x^{2}+1)^{1/2}$
$...$apply ln( ) to both sides
... on the RHS, $\ln(M\cdot N)=\ln M+\ln N$ and $\ln N^{r}=r\ln N$
$\displaystyle \ln y=\ln x+\frac{1}{2}\ln(x^{2}$+$1)\qquad $...$/\displaystyle \frac{d}{dx}$
... chain rule for the LHS and second term on the RHS
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{x}+\frac{1}{2}\cdot\frac{1}{x^{2}+1}\cdot 2x$
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{1}{x}+\frac{x}{x^{2}+1}$
$\displaystyle \frac{1}{y}(\frac{dy}{dx})=\frac{x^{2}+1+x^{2}}{x(x^{2}+1)} \qquad$...$/\times y$
$\displaystyle \frac{dy}{dx}=y[\frac{2x^{2}+1}{x(x^{2}+1)}]\qquad$
$... y=x\sqrt{x^{2}+1}=x(x^{2}$+$1)^{1/2}$
$\displaystyle \frac{dy}{dx}=\frac{x(x^{2}+1)^{1/2}(2x^{2}+1)}{x(x^{2}+1)}$
$\displaystyle \frac{dy}{dx}=\frac{2x^{2}+1}{(x^{2}+1)^{1/2}}$
