Q. 63 - 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.   

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

Step-by-Step Solution

Answer

The derivative of the function is $\frac{{\displaystyle -x^4-3x^2}}{{(x^3+5x^2-3x)}^2}$.

Step 1. Given Information

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

Step 2. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}\left(x^2\right)\times (x^3+5x^2-3x)-x^2{\displaystyle \frac{d}{dx}}(x^3+5x^2-3x)}{{(x^3+5x^2-3x)}^2}\\ & =& \frac{{\displaystyle \left(2x\right)(x^3+5x^2-3x)-x^2(3x^2+10x-3)}}{{(x^3+5x^2-3x)}^2}\\ & =& \frac{{\displaystyle (2x^4+10x^3-6x^2)-(3x^4+10x^3-3x^2)}}{{(x^3+5x^2-3x)}^2}\\ & =& \frac{{\displaystyle -x^4-3x^2}}{{(x^3+5x^2-3x)}^2}\end{array}$