Question

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

$f\left(x\right)={\left(2x+1\right)}^{3x}$

Short answer

The derivative of function is$y\text{'}={\left(2x+1\right)}^{3x}\left[\frac{6x}{2x+1}+\frac{3\ln (2x+1)}{x}\right]$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)={\left(2x+1\right)}^{3x}$

Step 2. Calculation    

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

$$\begin{gathered} \ln y=\ln {\left(2x+1\right)}^{3x} \\ \ln y=3x·\ln (2x+1) \\ \frac{1}{y}y\text{'}=\frac{3x}{2x+1}\left(2\right)+\frac{\ln (2x+1)}{x}\left(3\right) \\ y\text{'}={\left(2x+1\right)}^{3x}\left[\frac{6x}{2x+1}+\frac{3\ln (2x+1)}{x}\right] \end{gathered}$$