022 Sample Final A, Problem 11

Find the derivative: $g(x)={\frac {ln(x^{3}+7)}{(x^{4}+2x^{2})}}$ .

(Note: You do not need to simplify the derivative after finding it.)

Foundations:
This problem requires some more advanced rules of differentiation. In particular, it needs
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:

Find the derivative of the denominator:
We need to use the chain rule, where the inner function is $x^{3}+7$ and the outer function is natural log:
${\begin{array}{rcl}\left[\ln(x^{3}+7)\right]'&=&{\displaystyle {\frac {1}{x^{3}+7}}\cdot 3x^{2}}\\\\&=&{\displaystyle {\frac {3x^{2}}{x^{3}+7}}.}\end{array}}$

Apply the Quotient Rule:

${\begin{array}{rcl}\left[{\displaystyle {\frac {\ln(x^{3}+7)}{x^{4}+2x^{2}}}}\right]'&=&{\displaystyle {\frac {\left[\ln(x^{3}+7)\right]'\cdot \left(x^{4}+2x^{2}\right)-\left(x^{4}+2x^{2}\right)'\cdot \ln(x^{3}+7)}{\left(x^{4}+2x^{2}\right)^{2}}}}\\\\&=&{\displaystyle {\frac {{\frac {3x^{2}}{x^{3}+7}}\cdot \left(x^{4}+2x^{2}\right)-\left(4x^{3}+4x\right)\cdot \ln(x^{3}+7)}{\left(x^{4}+2x^{2}\right)^{2}}}}.\\\\\end{array}}$

Final Answer:
$\left[{\frac {\ln(x^{3}+7)}{x^{4}+2x^{2}}}\right]'\,=\,{\frac {{\frac {3x^{2}}{x^{3}+7}}\cdot \left(x^{4}+2x^{2}\right)-\left(4x^{3}+4x\right)\cdot \ln(x^{3}+7)}{\left(x^{4}+2x^{2}\right)^{2}}}.$

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