Q. 58 - Calculus

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.       

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

Step-by-Step Solution

Answer

The derivative of the function is $\frac{{\displaystyle (2x-3)(x^2-2x+1)-(x^2-3x)(2x-2)}}{{(x^2-2x+1)}^2}$.

Step 1. Given Information

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

Step 2. Find the derivative.

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

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