009A Sample Final A, Problem 2

2. Find the derivatives of the following functions:
   (a)  $f(x)={\frac {3x^{2}-5}{x^{3}-9}}.$

   (b)  $g(x)=\pi +2\cos({\sqrt {x-2}}).$

   (c)  $h(x)=4x\sin(x)+e(x^{2}+2)^{2}.$

Foundations:
These are problems involving some more advanced rules of differentiation. In particular, they use
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:

Part (a):
We need to use the quotient rule:
$f'(x)={\frac {\left(3x^{2}-5\right)'\cdot \left(x^{3}-9\right)-\left(3x^{2}-5\right)\cdot \left(x^{3}-9\right)'}{\left(x^{3}-9\right)^{2}}}$
$={\frac {6x(x^{3}-9)-(3x^{2}-5)(3x^{2})}{(x^{3}-9)^{2}}}$
$={\frac {6x^{4}-54x-9x^{4}+15x^{2}}{(x^{3}-9)^{2}}}$
$={\frac {-3x^{4}+15x^{2}-54x}{(x^{3}-9)^{2}}}.$

Part (b):
Both parts (b) and (c) attempt to confuse you by including the familiar constants $e$ and Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \pi } . Remember - they are just constants, like 10 or 1/2. With that in mind, we really just need to apply the chain rule to find
$g'(x)\,=\,0-2\sin \left({\sqrt {x-2}}\right)\cdot {\frac {1}{2}}\cdot (x-2)^{-1/2}\cdot 1\,=\,\,-\,{\frac {\sin \left({\sqrt {x-2}}\right)}{\sqrt {x-2}}}.$

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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 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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