Question

Use the differentiation rules developed in this section to find the derivatives of the functions in Exercises 35-64. Note that it may be necessary to do some preliminary algebra before differentiating.

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

Short answer

The derivative is$\begin{array}{rcl}& & \frac{{\displaystyle (2x-4)(x^3-3x^2+x-3)-(x^2+4-4x)(3x^2-6x+1)}}{{(x^3-3x^2+x-3)}^2}\end{array}$.

Step-by-step solution

Step 1. Given Information

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

Step 2. Simplify the function

Simplify the function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{x^2+4-4x}{x^2(x-3)+(x-3)}\\ & =& \frac{x^2+4-4x}{x^3-3x^2+x-3}\end{array}$

Step 3. Find the derivative

Apply the quotient rule of derivative, $\left(\frac{f}{g}\right)\text{'}\left(x\right)=\frac{f\text{'}\left(x\right)g\left(x\right)-f\left(x\right)g\text{'}\left(x\right)}{(g{\left(x\right))}^2}$.

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}(x^2+4-4x)\times (x^3-3x^2+x-3)-(x^2+4-4x){\displaystyle \frac{d}{dx}}(x^3-3x^2+x-3)}{{(x^3-3x^2+x-3)}^2}\\ & =& \frac{{\displaystyle (2x-4)(x^3-3x^2+x-3)-(x^2+4-4x)(3x^2-6x+1)}}{{(x^3-3x^2+x-3)}^2}\end{array}$