# 009A Sample Midterm 2, Problem 4

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

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

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

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

Solution:

(a)

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

(b)

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

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