009A Sample Midterm 3, Problem 6

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

(a)  $f(x)=\sin {\bigg (}{\frac {x^{-3}}{e^{-x}}}{\bigg )}$

b)  $g(x)={\sqrt {\frac {x^{2}+2}{x^{2}+4}}}$

c)  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 h(x)=(x+\cos^2x)^8}

Foundations:
1. Chain 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(g(x)))=f'(g(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}}

Solution:

(a)

Step 1:
First, using the Chain 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)=\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'.}

Step 2:
Now, using the Quotient Rule and Chain 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{x^{-3}}{e^{-x}}\bigg)'}\\ &&\\ & = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(x^{-3})'-x^{-3}(e^{-x})'}{(e^{-x})^2}\bigg)}\\ &&\\ & = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-x)'}{(e^{-x})^2}\bigg)}\\ &&\\ & = & \displaystyle{\cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg).} \end{array}}

(b)

Step 1:
First, using the Chain 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}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{x^2+2}{x^2+4}\bigg)'.}

Step 2:
Now, 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 \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{x^2+2}{x^2+4}\bigg)'}\\ &&\\ & = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(x^2+2)'-(x^2+2)(x^2+4)'}{(x^2+4)^2}\bigg)}\\ &&\\ & = & \displaystyle{\frac{1}{2}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg).} \end{array}}

(c)

Step 1:
First, using the Chain 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 h'(x)=8(x+\cos^2(x))^7(x+\cos^2(x))'.}

Step 2:
Now, using the Chain Rule again we 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{h'(x)} & = & \displaystyle{8(x+\cos^2(x))^7(x+\cos^2(x))'}\\ &&\\ & = & \displaystyle{8(x+\cos^2(x))^7(1+2\cos(x)(\cos(x))')}\\ &&\\ & = & \displaystyle{8(x+\cos^2(x))^7(1-2\cos(x)\sin(x)).} \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 \cos\bigg(\frac{x^{-3}}{e^{-x}}\bigg)\bigg(\frac{e^{-x}(-3x^{-4})-x^{-3}(e^{-x})(-1)}{(e^{-x})^2}\bigg)}
(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}\bigg(\frac{x^2+2}{x^2+4}\bigg)^{-\frac{1}{2}}\bigg(\frac{(x^2+4)(2x)-(x^2+2)(2x)}{(x^2+4)^2}\bigg)}
(c)     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 8(x+\cos^2(x))^7(1-2\cos(x)\sin(x))}

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