Q 37. - Calculus

Question

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

$f\left(x\right)=\ln \left(x^2+e^{\sqrt{x}}\right)$

Step-by-Step Solution

Answer

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

Step 1. Given information.

The given function is:

$f\left(x\right)=\ln \left(x^2+e^{\sqrt{x}}\right)$

Step 2. Find the derivative of the given function.

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