022 Exam 1 Sample A, Problem 8

8. Find the derivative of the function $f(x)={\frac {(3x-1)^{2}}{x^{3}-7}}$. 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 $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Quotient Rule: If $f$ and $g$ are differentiable functions and $g(x)\neq 0$ , then
$\left({\frac {f}{g}}\right)'(x)={\frac {f'(x)\cdot g(x)-f(x)\cdot g'(x)}{\left(g(x)\right)^{2}}}.$

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

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