007A Sample Midterm 2, Problem 3 Detailed Solution

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

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

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

(c)   $f(x)={\sqrt {3x^{2}+5x-7}}$  where  $x>0$

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Background Information:
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. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

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
$f'(x)={\frac {(1+6x)(x^{3}+x^{-3})'-(x^{3}+x^{-3})(1+6x)'}{(1+6x)^{2}}}.$

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

(c)

Step 1:
First, we write
$f(x)=(3x^{2}+5x-7)^{\frac {1}{2}}.$
Using the Chain Rule, we have
$f'(x)={\frac {1}{2}}(3x^{2}+5x-7)^{-{\frac {1}{2}}}(3x^{2}+5x-7)'.$

Step 2:
Now, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {{\frac {1}{2}}(3x^{2}+5x-7)^{-{\frac {1}{2}}}(3x^{2}+5x-7)'}\\&&\\&=&\displaystyle {{\frac {1}{2}}(3x^{2}+5x-7)^{-{\frac {1}{2}}}(6x+5).}\end{array}}$

Final Answer:
(a)     $f'(x)=(x^{3}){\bigg (}{\frac {4}{3}}x^{\frac {1}{3}}{\bigg )}+(3x^{2})(x^{\frac {4}{3}}-1)$
(b)     $f'(x)={\frac {(1+6x)(3x^{2}-3x^{-4})-(x^{3}+x^{-3})(6)}{(1+6x)^{2}}}$
(c)     $f'(x)={\frac {1}{2}}(3x^{2}+5x-7)^{-{\frac {1}{2}}}(6x+5)$

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