let $y = sqrt{xe^{x}}$. $\frac{dy}{dx}=$

Explanation:

Step1: Apply product - rule

The product - rule states that if $y = uv$, where $u$ and $v$ are functions of $x$, then $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$. Here, let $u = \sqrt{x}$ and $v = e^{x}$.

Step2: Find $\frac{du}{dx}$

We know that if $u=\sqrt{x}=x^{\frac{1}{2}}$, then by the power - rule $\frac{du}{dx}=\frac{1}{2}x^{-\frac{1}{2}}=\frac{1}{2\sqrt{x}}$.

Step3: Find $\frac{dv}{dx}$

Since $v = e^{x}$, then $\frac{dv}{dx}=e^{x}$.

Step4: Calculate $\frac{dy}{dx}$

Using the product - rule $\frac{dy}{dx}=u\frac{dv}{dx}+v\frac{du}{dx}$, we substitute $u = \sqrt{x}$, $v = e^{x}$, $\frac{du}{dx}=\frac{1}{2\sqrt{x}}$ and $\frac{dv}{dx}=e^{x}$: [ \begin{align*} \frac{dy}{dx}&=\sqrt{x}\cdot e^{x}+e^{x}\cdot\frac{1}{2\sqrt{x}}\ &=e^{x}\left(\sqrt{x}+\frac{1}{2\sqrt{x}}\right)\ &=\frac{e^{x}(2x + 1)}{2\sqrt{x}} \end{align*} ]

Answer:

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