# 007A Sample Midterm 3, Problem 3 Detailed Solution

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

(a)  ${\displaystyle f(x)={\frac {(3x-5)(-x^{-2}+4x)}{x^{\frac {4}{5}}}}}$

(b)  ${\displaystyle g(x)={\sqrt {x}}+{\frac {1}{\sqrt {x}}}+{\sqrt {\pi }}}$  for  ${\displaystyle x>0.}$

(c)  ${\displaystyle h(x)={\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{4}}$

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

Solution:

(a)

Step 1:
Using the Quotient Rule, we have
${\displaystyle f'(x)={\frac {x^{\frac {4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac {4}{5}})'}{(x^{\frac {4}{5}})^{2}}}.}$
Step 2:
Now, we use the Product Rule and Power Rule to get

${\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{\frac {4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac {4}{5}})'}{(x^{\frac {4}{5}})^{2}}}\\&&\\&=&\displaystyle {\frac {x^{\frac {4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{-{\frac {1}{5}}})}{(x^{\frac {4}{5}})^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{\frac {4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{-{\frac {1}{5}}})}{(x^{\frac {4}{5}})^{2}}}.}\end{array}}}$

(b)

Step 1:
First, we have
${\displaystyle g'(x)=({\sqrt {x}})'+{\bigg (}{\frac {1}{\sqrt {x}}}{\bigg )}'+({\sqrt {\pi }})'.}$
Step 2:
Since  ${\displaystyle \pi }$  is a constant,  ${\displaystyle {\sqrt {\pi }}}$  is also a constant.
Hence,
${\displaystyle ({\sqrt {\pi }})'=0.}$
Therefore, using the Power Rule, we have
${\displaystyle {\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {({\sqrt {x}})'+{\bigg (}{\frac {1}{\sqrt {x}}}{\bigg )}'+({\sqrt {\pi }})'}\\&&\\&=&\displaystyle {{\frac {1}{2}}x^{-{\frac {1}{2}}}+-{\frac {1}{2}}x^{-{\frac {3}{2}}}+0}\\&&\\&=&\displaystyle {{\frac {1}{2}}x^{-{\frac {1}{2}}}+-{\frac {1}{2}}x^{-{\frac {3}{2}}}.}\end{array}}}$

(c)

Step 1:
First, using the Chain Rule, we have
${\displaystyle h'(x)=4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}'.}$
Step 2:
Now, using the Quotient Rule and Power Rule, we get
${\displaystyle {\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}'}\\&&\\&=&\displaystyle {4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {(x+1)(3x^{2})'-(3x^{2})(x+1)'}{(x+1)^{2}}}{\bigg )}}\\&&\\&=&\displaystyle {4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {(x+1)(6x)-3x^{2}}{(x+1)^{2}}}{\bigg )}.}\end{array}}}$

(a)     ${\displaystyle f'(x)={\frac {x^{\frac {4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{-{\frac {1}{5}}})}{(x^{\frac {4}{5}})^{2}}}}$
(b)     ${\displaystyle g'(x)={\frac {1}{2}}x^{-{\frac {1}{2}}}+-{\frac {1}{2}}x^{-{\frac {3}{2}}}}$
(c)     ${\displaystyle h'(x)=4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {(x+1)(6x)-3x^{2}}{(x+1)^{2}}}{\bigg )}}$