# Difference between revisions of "022 Exam 1 Sample A, Problem 8"

8. Find the derivative of the function ${\displaystyle f(x)={\frac {(3x-1)^{2}}{x^{3}-7}}}$. You do not need to simplify your answer.

Foundations:
This problem involves some more advanced rules of differentiation. In particular, it requires
The Chain Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions, then

${\displaystyle (f\circ g)'(x)=f'(g(x))\cdot g'(x).}$

The Quotient Rule: If ${\displaystyle f}$ and ${\displaystyle g}$ are differentiable functions and ${\displaystyle g(x)\neq 0}$ , then

${\displaystyle \left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.}$

Solution:
Note that we need to use chain rule to find the derivative of ${\displaystyle \left(3x-1\right)^{2}}$. Then we find
 ${\displaystyle f'(x)}$ ${\displaystyle =\,\,{\frac {\left[\left(3x-1\right)^{2}\right]'\cdot (x^{3}-7)\,\,-\,\,\left(3x-1\right)^{2}\cdot (x^{3}-7)'}{(x^{3}-7)^{2}}}}$ ${\displaystyle =\,\,{\frac {\left[2\left(3x-1\right)\cdot 3\right]\cdot (x^{3}-7)\,\,-\,\,\left(3x-1\right)^{2}\cdot 3x^{2}}{(x^{3}-7)^{2}}}.}$