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.

$f\left(x\right)=\frac{x^7-3x^5+4}{1-3x^4}$

Short answer

The derivative of the function is$f\text{'}\left(x\right)=\frac{{\displaystyle (7x^6-15x^4)(1-3x^4)-(x^7-3x^5+4)(-12x^3)}}{{(1-3x^4)}^2}$.

Step-by-step solution

Step 1. Given Information

The given function is$f\left(x\right)=\frac{x^7-3x^5+4}{1-3x^4}$.

Step 2. Find the derivative

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

$\begin{array}{rcl}f\text{'}\left(x\right)& =& \frac{{\displaystyle \frac{d}{dx}}(x^7-3x^5+4)·(1-3x^4)-(x^7-3x^5+4){\displaystyle \frac{d}{dx}}(1-3x^4)}{{(1-3x^4)}^2}\\ & =& \frac{{\displaystyle (7x^6-15x^4)(1-3x^4)-(x^7-3x^5+4)(-12x^3)}}{{(1-3x^4)}^2}\end{array}$