007A Sample Midterm 3, Problem 3 Detailed Solution

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

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

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

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

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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. Power Rule
${\frac {d}{dx}}(x^{n})=nx^{n-1}$
4. Chain Rule
${\frac {d}{dx}}(f(g(x)))=f'(g(x))g'(x)$

Solution:

(a)

Step 1:
Using the Quotient Rule, we have
$f'(x)={\frac {x^{\frac {4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac {4}{5}})'}{(x^{\frac {4}{5}})^{2}}}.$

Step 2:
Now, we use the Product Rule to get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{\frac {4}{5}}((3x-5)(-x^{-2}+4x))'-(3x-5)(-x^{-2}+4x)(x^{\frac {4}{5}})'}{(x^{\frac {4}{5}})^{2}}}\\&&\\&=&\displaystyle {\frac {x^{\frac {4}{5}}[(3x-5)(-x^{-2}+4x)'+(3x-5)'(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{-{\frac {1}{5}}})}{(x^{\frac {4}{5}})^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{\frac {4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{-{\frac {1}{5}}})}{(x^{\frac {4}{5}})^{2}}}.}\end{array}}$

(b)

Step 1:
First, we have
$g'(x)=({\sqrt {x}})'+{\bigg (}{\frac {1}{\sqrt {x}}}{\bigg )}'+({\sqrt {\pi }})'.$

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

(c)

Step 1:
First, using the Chain Rule, we have
$h'(x)=4{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}^{3}{\bigg (}{\frac {3x^{2}}{x+1}}{\bigg )}'.$

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

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

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