009A Sample Midterm 3, Problem 5

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 $x>0.$

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 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
${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {({\sqrt {x}})'+{\bigg (}{\frac {1}{\sqrt {x}}}{\bigg )}'+({\sqrt {\pi }})'}\\&&\\&=&\displaystyle {{\frac {1}{2}}x^{\frac {-1}{2}}+{\frac {-1}{2}}x^{\frac {-3}{2}}+0}\\&&\\&=&\displaystyle {{\frac {1}{2}}x^{\frac {-1}{2}}+{\frac {-1}{2}}x^{\frac {-3}{2}}.}\end{array}}$

Final Answer:
(a)     ${\frac {x^{\frac {4}{5}}[(3x-5)(2x^{-3}+4)+(3)(-x^{-2}+4x)]-(3x-5)(-x^{-2}+4x)({\frac {4}{5}}x^{\frac {-1}{5}})}{(x^{\frac {4}{5}})^{2}}}$
(b)     ${\frac {1}{2}}x^{\frac {-1}{2}}+{\frac {-1}{2}}x^{\frac {-3}{2}}$

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