009A Sample Midterm 3, Problem 6

Find the derivatives of the following functions. Do not simplify.

a)$f(x)=\sin {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}$

b)$g(x)={\sqrt {\frac {x^{2}+2}{x^{2}+4}}}$

c)$h(x)=(x+\cos ^{2}x)^{8}$

Foundations:
1. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$
2. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$

Solution:

(a)

Step 1:
First, using the Chain Rule, we have
$f'(x)=\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}'.$

Step 2:
Now, using the Quotient Rule and Chain Rule, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}'}\\&&\\&=&\displaystyle {\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {e^{-x}(x^{-3})'-x^{-3}(e^{-x})'}{(e^{-x})^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-x)'}{(e^{-x})^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^{2}}}{\bigg )}.}\end{array}}$

(b)

Step 1:
First, using the Chain Rule, we have
$g'(x)={\frac {1}{2}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}^{-{\frac {1}{2}}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}'.$

Step 2:
Now, using the Quotient Rule, we have
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {{\frac {1}{2}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}^{-{\frac {1}{2}}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}'}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}^{-{\frac {1}{2}}}{\bigg (}{\frac {(x^{2}+4)(x^{2}+2)'-(x^{2}+2)(x^{2}+4)'}{(x^{2}+4)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {{\frac {1}{2}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}^{-{\frac {1}{2}}}{\bigg (}{\frac {(x^{2}+4)(2x)-(x^{2}+2)(2x)}{(x^{2}+4)^{2}}}{\bigg )}.}\end{array}}$

(c)

Step 1:
First, using the Chain Rule, we have
$h'(x)=8(x+\cos ^{2}(x))^{7}(x+\cos ^{2}(x))'.$

Step 2:
Now, using the Chain Rule again we get
${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {8(x+\cos ^{2}(x))^{7}(x+\cos ^{2}(x))'}\\&&\\&=&\displaystyle {8(x+\cos ^{2}(x))^{7}(1+2\cos(x)(\cos(x))')}\\&&\\&=&\displaystyle {8(x+\cos ^{2}(x))^{7}(1-2\cos(x)\sin(x)).}\end{array}}$

Final Answer:
(a)     $\cos {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}{\bigg (}{\frac {e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^{2}}}{\bigg )}$
(b)     ${\frac {1}{2}}{\bigg (}{\frac {x^{2}+2}{x^{2}+4}}{\bigg )}^{-{\frac {1}{2}}}{\bigg (}{\frac {(x^{2}+4)(2x)-(x^{2}+2)(2x)}{(x^{2}+4)^{2}}}{\bigg )}$
(c)     $8(x+\cos ^{2}(x))^{7}(1-2\cos(x)\sin(x))$

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