022 Exam 2 Sample A, Problem 1

Find the derivative of  $y\,=\,\ln {\frac {(x+5)(x-1)}{x}}.$

Foundations:
This problem requires several advanced rules of differentiation. In particular, you need
The Chain Rule: If $f$ and $g$ are differentiable functions, then
$(f\circ g)'(x)=f'(g(x))\cdot g'(x).$
The Product Rule: If $f$ and $g$ are differentiable functions, then
$(fg)'(x)=f'(x)\cdot g(x)+f(x)\cdot g'(x).$
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:

Step 1:
We need to identify the composed functions in order to apply the chain rule. Note that if we set $g(x)\,=\,\ln x$, and
$f(x)\,=\,{\frac {(x+5)(x-1)}{x}},$
we then have $y\,=\,g\circ f(x)\,=\,g\left(f(x)\right).$

Step 2:
We can now apply all three advanced techniques. For example, to find the derivative $f'(x)$,

Part (c):
We can choose to expand the second term, finding
$e(x^{2}+2)^{2}=ex^{4}+4ex^{2}+4e.$
We then only require the product rule on the first term, so
$h'(x)\,=\,(4x)'\cdot \sin(x)+4x\cdot (\sin(x))'+\left(ex^{4}+4ex^{2}+4e\right)'\,=\,4\sin(x)+4x\cos(x)+4ex^{3}+8ex.$

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