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.

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

Short answer

The derivative of the function is$\begin{array}{rcl}& & \frac{{\displaystyle 5-2x}}{{(x^2-5x+6)}^2}\end{array}$.

Step-by-step solution

Step 1. Given Information

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

Step 2. Simplify the function

Simplify the denominator of the given function.

$\begin{array}{rcl}f\left(x\right)& =& \frac{1}{x(x-3)-2(x-3)}\\ & =& \frac{1}{x^2-3x-2x+6}\\ & =& \frac{1}{x^2-5x+6}\end{array}$

Step 3. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}\left(1\right)\times (x^2-5x+6)-1{\displaystyle \frac{d}{dx}}(x^2-5x+6)}{{(x^2-5x+6)}^2}\\ & =& \frac{{\displaystyle \left(0\right)\times (x^2-5x+6)-(2x-5)}}{{(x^2-5x+6)}^2}\\ & =& \frac{{\displaystyle 5-2x}}{{(x^2-5x+6)}^2}\end{array}$