009A Sample Midterm 2, Problem 4

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

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

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

Foundations:
1. Product Rule
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{d}{dx}(f(x)g(x))=f(x)g'(x)+f'(x)g(x)}
2. Quotient Rule
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{d}{dx}\bigg(\frac{f(x)}{g(x)}\bigg)=\frac{g(x)f'(x)-f(x)g'(x)}{(g(x))^2}}
3. Power Rule
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{d}{dx}(x^n)=nx^{n-1}}

Solution:

(a)

Step 1:
Using the Product 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)=x^3(x^{\frac{4}{3}}-1)'+(x^3)'(x^{\frac{4}{3}}-1).}

Step 2:
Now, 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{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
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)=\frac{(1+6x)(x^3+x^{-3})'-(x^3+x^{-3})(1+6x)'}{(1+6x)^2}.}

Step 2:
Now, 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{\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}}

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 x^3\bigg(\frac{4}{3}x^{\frac{1}{3}}\bigg)+(3x^2)(x^{\frac{4}{3}}-1)}
(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+6x)(3x^2-3x^{-4})-(x^3+x^{-3})(6)}{(1+6x)^2}}

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