Difference between revisions of "009A Sample Final 1, Problem 3"
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|Using the Chain Rule, we have | |Using the Chain Rule, we have | ||
|- | |- | ||
| − | | | + | | |
| − | + | <math>\begin{array}{rcl} | |
\displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ | \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ | ||
&&\\ | &&\\ | ||
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|To do this, we use the Quotient Rule. So, we have | |To do this, we use the Quotient Rule. So, we have | ||
|- | |- | ||
| − | | | + | | |
| − | + | <math>\begin{array}{rcl} | |
\displaystyle{f'(x)} & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ | \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ | ||
&&\\ | &&\\ | ||
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!Step 1: | !Step 1: | ||
|- | |- | ||
| − | | | + | |We need to use the Chain Rule. We have |
|- | |- | ||
| | | | ||
| − | + | <math>g'(x)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).</math> | |
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |We need to calculate& | + | |We need to calculate <math>\frac{d}{dx}\sqrt{1+x^3}.</math> |
|- | |- | ||
|We use the Chain Rule again to get | |We use the Chain Rule again to get | ||
|- | |- | ||
| | | | ||
| − | + | <math>\begin{array}{rcl} | |
\displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\ | \displaystyle{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\ | ||
&&\\ | &&\\ | ||
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!Final Answer: | !Final Answer: | ||
|- | |- | ||
| − | |'''(a)''' <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math> | + | | '''(a)''' <math style="vertical-align: -14px">f'(x)=\frac{4x}{x^4-1}</math> |
|- | |- | ||
| − | |'''(b)''' <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math> | + | | '''(b)''' <math style="vertical-align: -18px">g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}</math> |
|} | |} | ||
[[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | ||
Revision as of 16:38, 25 February 2017
Find the derivatives of the following functions.
(a)
(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 g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})}
| 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}} |
| 3. Trig Derivatives |
| 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}(\sin x)=\cos x,\quad\frac{d}{dx}(\tan x)=\sec^2 x} |
Solution:
(a)
| Step 1: |
|---|
| 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 \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{\bigg(\frac{x^2-1}{x^2+1}\bigg)}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).}\\ \end{array}} |
| Step 2: |
|---|
| Now, we need to calculate 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 \bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).} |
| To do this, we use the Quotient Rule. So, 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{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{(x^2+1)(2x)-(x^2-1)(2x)}{(x^2+1)^2}\bigg)}\\ &&\\ & = & \displaystyle{\frac{x^2+1}{x^2-1}\bigg(\frac{4x}{(x^2+1)^2}\bigg)}\\ &&\\ & = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\ &&\\ & = & \displaystyle{\frac{4x}{x^4-1}.}\\ \end{array}} |
(b)
| Step 1: |
|---|
| We need to use 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)\,=\,8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg).} |
| Step 2: |
|---|
| We need to calculate 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}\sqrt{1+x^3}.} |
| We use the Chain Rule again 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{g'(x)} & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)}\\ &&\\ & = & \displaystyle{8\cos(4x)+4\sec^2(\sqrt{1+x^3})\frac{1}{2}(1+x^3)^{-\frac{1}{2}}3x^2}\\ &&\\ & = & \displaystyle{8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}.}\\ \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 f'(x)=\frac{4x}{x^4-1}} |
| (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 g'(x)=8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}} |