Question

Find the derivatives of the functions:$f\left(x\right)=\ln \left(\frac{x^3}{x^2+x+1}\right)$

Short answer

The required answer is$\frac{x^2+2x+3}{x(x^2+x+1)}$.

Step-by-step solution

Step 1. Given information. 

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

Step 2. Find the derivative of the given function.

$$\begin{gathered} f\left(x\right)=\ln \left(\frac{x^3}{x^2+x+1}\right) \\ f\text{'}\left(x\right)=\frac{d}{dx}\left[\ln \left(\frac{x^3}{x^2+x+1}\right)\right] \\ =\frac{1}{\frac{x^3}{x^2+x+1}}\frac{d}{dx}\left(\frac{x^3}{x^2+x+1}\right) \\ =\frac{x^2+x+1}{x^3}\frac{\left(x^2+x+1\right){\displaystyle \frac{d}{dx}}\left(x^3\right)-x^3\frac{d}{dx}\left(x^2+x+1\right)}{{\left(x^2+x+1\right)}^2} \\ =\frac{x^2+x+1}{x^3}\frac{3x^2\left(x^2+x+1\right)-x^3(2x+1)}{{\left(x^2+x+1\right)}^2} \\ =\frac{3x^4+3x^3+3x^2-2x^4-x^3}{x^3\left(x^2+x+1\right)} \\ =\frac{x^4+2x^3+3x^2}{x^3\left(x^2+x+1\right)} \\ =\frac{x^2\left(x^2+2x+3\right)}{x^3\left(x^2+x+1\right)} \\ =\frac{\left(x^2+2x+3\right)}{x\left(x^2+x+1\right)} \end{gathered}$$