Derivative of sqrtx*logexp

1. Apply the product rule:

   $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

   $f{\left(x \right)} = \sqrt{x}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Apply the power rule: $\sqrt{x}$ goes to $\frac{1}{2 \sqrt{x}}$

   $g{\left(x \right)} = \log{\left(e^{x} \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = e^{x}$.

   2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$.

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} e^{x}$:

      1. The derivative of $e^{x}$ is itself.

      The result of the chain rule is:

      $$1$$

   The result is: $\sqrt{x} + \frac{\log{\left(e^{x} \right)}}{2 \sqrt{x}}$

2. Now simplify:

   $$\frac{3 \sqrt{x}}{2}$$

The answer is:

$$\frac{3 \sqrt{x}}{2}$$