Question

Find the derivatives of each of functions in Exercises 17–44. In some cases it may be convenient to do some preliminary algebra.

$f\left(x\right)=\frac{\sqrt{e^x}}{\ln \sqrt{x}}$

Short answer

The required answer is$\frac{\left(\ln \sqrt{x}·\sqrt{e^x}\right)·{\displaystyle \frac{1}{2}}-\sqrt{e^x}·{\displaystyle \frac{1}{2x}}}{{\left(\ln \sqrt{x}\right)}^2}$.

Step-by-step solution

Step 1. Given information.

The given function is:

$f\left(x\right)=\frac{\sqrt{e^x}}{\ln \sqrt{x}}$

Step 2. Find the derivative of the given function.

$$\begin{gathered} f\left(x\right)=\frac{\sqrt{e^x}}{\ln \sqrt{x}} \\ f\text{'}\left(x\right)=\frac{d}{dx}\frac{\sqrt{e^x}}{\ln \sqrt{x}} \\ =\frac{\ln \sqrt{x}{\displaystyle \frac{d}{dx}}\sqrt{e^x}-\sqrt{e^x}·{\displaystyle \frac{d}{dx}}\ln \sqrt{x}}{{\left(\ln \sqrt{x}\right)}^2} \\ =\frac{\ln \sqrt{x}·{\displaystyle \frac{1}{2\sqrt{e^x}}}·e^x-\sqrt{e^x}·{\displaystyle \frac{1}{\sqrt{x}}}·{\displaystyle \frac{1}{2\sqrt{x}}}}{{\left(\ln \sqrt{x}\right)}^2} \\ =\frac{\left(\ln \sqrt{x}·\sqrt{e^x}\right)·{\displaystyle \frac{1}{2}}-\sqrt{e^x}·{\displaystyle \frac{1}{2x}}}{{\left(\ln \sqrt{x}\right)}^2} \end{gathered}$$