009A Sample Midterm 2, Problem 4

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

a) $f(x)=x^{3}(x^{\frac {4}{3}}-1)$

b) $g(x)={\frac {x^{3}+x^{-3}}{1+6x}}$ where $x>0$

Foundations:
1. Product Rule
${\frac {d}{dx}}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)$
2. Quotient Rule
${\frac {d}{dx}}{\bigg (}{\frac {f(x)}{g(x)}}{\bigg )}={\frac {g(x)f'(x)-f(x)g'(x)}{(g(x))^{2}}}$
3. Power Rule
${\frac {d}{dx}}(x^{n})=nx^{n-1}$

Solution:

(a)

Step 1:
Using the Product Rule, we have
$f'(x)=x^{3}(x^{\frac {4}{3}}-1)'+(x^{3})'(x^{\frac {4}{3}}-1).$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {x^{3}(x^{\frac {4}{3}}-1)'+(x^{3})'(x^{\frac {4}{3}}-1)}\\&&\\&=&\displaystyle {x^{3}{\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1).}\end{array}}$

(b)

Step 1:
Using the Quotient Rule, we have
$g'(x)={\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}\\&&\\&=&\displaystyle {{\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}.}\end{array}}$

Final Answer:
(a)     $x^{3}{\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1)$
(b)     ${\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}$

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