Difference between revisions of "009A Sample Final 1, Problem 3"
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\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)}\\ | ||
&&\\ | &&\\ | ||
| − | & = & \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{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).}\\ |
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
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!Step 2: | !Step 2: | ||
|- | |- | ||
| − | |Now, we need to calculate <math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)</math> | + | |Now, we need to calculate <math>\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg).</math> |
|- | |- | ||
|To do this, we use the Quotient Rule. So, we have | |To do this, we use the Quotient Rule. So, we have | ||
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& = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\ | & = & \displaystyle{\frac{4x}{(x^2-1)(x^2+1)}}\\ | ||
&&\\ | &&\\ | ||
| − | & = & \displaystyle{\frac{4x}{x^4-1}}\\ | + | & = & \displaystyle{\frac{4x}{x^4-1}.}\\ |
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
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|- | |- | ||
| | | | ||
| − | ::<math>g'(x)=8\cos(4x)+4\sec^2(\sqrt{1+x^3})\bigg(\frac{d}{dx}\sqrt{1+x^3}\bigg)</math> | + | ::<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 <math>\frac{d}{dx}\sqrt{1+x^3}</math> | + | |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 | ||
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& = & \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)+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}}}\\ | + | & = & \displaystyle{8\cos(4x)+\frac{6\sec^2(\sqrt{1+x^3})x^2}{\sqrt{1+x^3}}.}\\ |
\end{array}</math> | \end{array}</math> | ||
|} | |} | ||
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|'''(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 12:29, 29 February 2016
Find the derivatives of the following functions.
a)
b)
| Foundations: |
|---|
| For functions , recall |
| Chain Rule |
| Quotient Rule |
| Trig derivatives |
Solution:
(a)
| Step 1: |
|---|
| Using the Chain Rule, we have |
|
|
| Step 2: |
|---|
| Now, we need to calculate |
| To do this, we use the Quotient Rule. So, we have |
|
|
(b)
| Step 1: |
|---|
| Again, we need to use the Chain Rule. We have |
|
|
| Step 2: |
|---|
| We need to calculate |
| We use the Chain Rule again to get |
|
|
| Final Answer: |
|---|
| (a) |
| (b) |
