Q 40. - 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+1\right){\left(e^x\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Step-by-Step Solution

Answer

The required answer is $f\text{'}\left(x\right)=-\frac{1}{3}{e}^x \ln \left(x^2+1\right)+{\left(e^x\right)}^{-\frac{1}{3}}·\frac{2x}{x^2+1}$

Step 1. Given information.

The given function is:

$f\left(x\right)=\ln \left(x^2+1\right){\left(e^x\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Step 2. Find the derivative of the given function.

$$\begin{gathered} f\left(x\right)=\ln \left(x^2+1\right){\left(e^x\right)}^{-\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.} \\ f\text{'}\left(x\right)=\frac{d}{dx}\left(\ln \left(x^2+1\right){\left(e^x\right)}^{-\frac{1}{3}}\right) \\ =-\frac{1}{3}{e}^x \ln \left(x^2+1\right)+{\left(e^x\right)}^{-\frac{1}{3}}·\frac{1}{x^2+1}·2x \\ =-\frac{1}{3}{e}^x \ln \left(x^2+1\right)+{\left(e^x\right)}^{-\frac{1}{3}}·\frac{2x}{x^2+1} \end{gathered}$$