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 Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle x>0.}

Foundations:
1. Quotient Rule
2. Product Rule
3. Power Rule

Solution:

(a)

Step 1:
Using the Quotient Rule, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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 to get
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=(\sqrt{x})'+\bigg(\frac{1}{\sqrt{x}}\bigg)'+(\sqrt{\pi})'.}

Step 2:
Since Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \pi} is a constant, Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sqrt{\pi}} is also a constant.
Hence,
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\sqrt{\pi})'=0.}
Therefore, we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}

Final Answer:
(a)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\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}}
(b)     Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{1}{2}x^{\frac{-1}{2}}+\frac{-1}{2}x^{\frac{-3}{2}}}

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