# 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
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}\\&&\\&=&\displaystyle {{\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}.}\end{array}}$ (a)     $f'(x)=x^{3}{\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1)$ (b)     $g'(x)={\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}$ 