Q. 59 - 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.        

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

Step-by-Step Solution

Answer

The derivative is $\begin{array}{rcl}& & \frac{23}{{(5x+4)}^2}\end{array}$.

Step 1. Given Information

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

Step 2. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}(2x-3)\times (5x+4)-(2x-3){\displaystyle \frac{d}{dx}}(5x+4)}{{(5x+4)}^2}\\ & =& \frac{{\displaystyle \left(2\right)(5x+4)-(2x-3)\left(5\right)}}{{(5x+4)}^2}\\ & =& \frac{{\displaystyle (10x+8)-(10x-15)}}{{(5x+4)}^2}\\ & =& \frac{23}{{(5x+4)}^2}\end{array}$