|
|
| (17 intermediate revisions by 2 users not shown) |
| Line 1: |
Line 1: |
| | <span class="exam">Find the derivatives of the following functions. | | <span class="exam">Find the derivatives of the following functions. |
| | | | |
| − | <span class="exam">a) <math style="vertical-align: -14px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math> | + | <span class="exam">(a) <math style="vertical-align: -14px">f(x)=\ln \bigg(\frac{x^2-1}{x^2+1}\bigg)</math> |
| | | | |
| − | <span class="exam">b) <math style="vertical-align: -3px">g(x)=2\sin (4x)+4\tan (\sqrt{1+x^3})</math> | + | <span class="exam">(b) <math style="vertical-align: -3px">g(x)=3\sin (4x)+4\tan (\sqrt{1+x^3})</math> |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| + | <hr> |
| − | !Foundations:
| + | [[009A Sample Final 1, Problem 3 Solution|'''<u>Solution</u>''']] |
| − | |-
| |
| − | |'''1.''' Chain Rule
| |
| − | |-
| |
| − | |'''2.''' Quotient rule
| |
| − | |-
| |
| − | |'''3.''' derivatives of trig functions | |
| − | |}
| |
| | | | |
| − | '''Solution:'''
| |
| | | | |
| − | '''(a)''' | + | [[009A Sample Final 1, Problem 3 Detailed Solution|'''<u>Detailed Solution</u>''']] |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |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{\frac{x^2+1}{x^2-1}\bigg(\frac{d}{dx}\bigg(\frac{x^2-1}{x^2+1}\bigg)\bigg)}\\
| |
| − | \end{array}</math>
| |
| − | |}
| |
| | | | |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |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
| |
| − | |-
| |
| − | |
| |
| − | ::<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{\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}</math>
| |
| − | |}
| |
| − |
| |
| − | '''(b)'''
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 1:
| |
| − | |-
| |
| − | |Again, 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>.
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Step 2:
| |
| − | |-
| |
| − | |We need to calculate <math>\frac{d}{dx}\sqrt{1+x^3}</math>.
| |
| − | |-
| |
| − | |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{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}</math>
| |
| − | |}
| |
| − |
| |
| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
| |
| − | !Final Answer:
| |
| − | |-
| |
| − | |'''(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>.
| |
| − | |}
| |
| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
