Question

Use logarithmic differentiation to find the derivatives of each of the functions in Exercises 49–58.

$f\left(x\right)={\left(\frac{1}{x+1}\right)}^x$

Short answer

The derivative of function is$y\text{'}={\left(\frac{1}{x+1}\right)}^x\left[\frac{-x}{{\displaystyle x+1}}+\ln \left(\frac{1}{x+1}\right)\right]$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)={\left(\frac{1}{x+1}\right)}^x$

Step 2. Calculation    

First, we will take the log on both sides and then differentiate it,

$$\begin{gathered} \ln y=\ln {\left(\frac{1}{x+1}\right)}^x \\ \ln y=x\ln \left(\frac{1}{x+1}\right) \\ \frac{1}{y}y\text{'}=x\frac{1}{{\displaystyle \frac{1}{x+1}}}\left(\frac{(x+1)0-1}{{(x+1)}^2}\right)+\ln \left(\frac{1}{x+1}\right) \\ y\text{'}={\left(\frac{1}{x+1}\right)}^x\left[\frac{-x}{{\displaystyle x+1}}+\ln \left(\frac{1}{x+1}\right)\right] \end{gathered}$$