Question

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

$f\left(x\right)=\frac{e^{2x}{\left(x^3-2\right)}^4}{x\left(3e^{5x}+1\right)}$

Short answer

The derivative of function is$y\text{'}=\frac{e^{2x}{\left(x^3-2\right)}^4}{x\left(3e^{5x}+1\right)}\left[2+\frac{12x^2}{x^3-2}-\frac{1}{x}-\frac{5e^{5x}}{3e^{5x}+1}\right]$

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=\frac{e^{2x}{\left(x^3-2\right)}^4}{x\left(3e^{5x}+1\right)}$

Step 2. Calculation    

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

$$\begin{gathered} \ln y=\ln \left[e^{2x}{\left(x^3-2\right)}^4\right]-\ln \left(x\left(3e^{5x}+1\right)\right) \\ \ln y=\ln \left(e^{2x}\right)+\ln {\left(x^3-2\right)}^4-\ln x-\ln \left(3e^{5x}+1\right) \\ \ln y=2x+4\ln \left(x^3-2\right)-\ln x-\ln \left(3e^{5x}+1\right) \\ \frac{1}{y}y\text{'}=2+\frac{4}{x^3-2}\left(3x^2\right)-\frac{1}{x}-\frac{1}{3e^{5x}+1}\left(5e^{5x}\right) \\ y\text{'}=\frac{e^{2x}{\left(x^3-2\right)}^4}{x\left(3e^{5x}+1\right)}\left[2+\frac{12x^2}{x^3-2}-\frac{1}{x}-\frac{5e^{5x}}{3e^{5x}+1}\right] \end{gathered}$$