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| − | <span class="exam"> If | + | <span class="exam"> If <math style="vertical-align: -5px">y=\cos^{-1} (2x)</math> compute <math style="vertical-align: -12px">\frac{dy}{dx}</math> and find the equation for the tangent line at <math style="vertical-align: -14px">x_0=\frac{\sqrt{3}}{4}.</math> |
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| − | ::::::<math>y=x^2+\cos (\pi(x^2+1))</math>
| + | <span class="exam">You may leave your answers in point-slope form. |
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| − | <span class="exam">compute  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  and find the equation for the tangent line at <math style="vertical-align: -3px">x_0=1</math>. You may leave your answers in point-slope form. | + | <hr> |
| | + | [[009A Sample Final 1, Problem 4 Solution|'''<u>Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Foundations:
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| − | |'''1.''' What two pieces of information do you need to write the equation of a line?
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| − | ::You need the slope of the line and a point on the line.
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| − | |'''2.''' What does the Chain Rule state?
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| − | ::For functions <math style="vertical-align: -5px">f(x)</math>  and <math style="vertical-align: -5px">g(x),</math> <math style="vertical-align: -12px">~\frac{d}{dx}(f(g(x)))=f'(g(x))g'(x).</math>
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| − | |}
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| − | '''Solution:''' | + | [[009A Sample Final 1, Problem 4 Detailed Solution|'''<u>Detailed Solution</u>''']] |
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 1:
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| − | |First, we compute  <math style="vertical-align: -13px">\frac{dy}{dx}.</math> We get
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| − | ::<math>\frac{dy}{dx}\,=\,2x-\sin(\pi(x^2+1))(2\pi x).</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Step 2:
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| − | |To find the equation of the tangent line, we first find the slope of the line.
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| − | |Using <math style="vertical-align: -3px">x_0=1</math>  in the formula for  <math style="vertical-align: -12px">\frac{dy}{dx}</math>  from Step 1, we get
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| − | ::<math>m=2(1)-\sin(2\pi)2\pi\,=\,2.</math>
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| − | |To get a point on the line, we plug in <math style="vertical-align: -3px">x_0=1</math>  into the equation given.
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| − | |So, we have  <math style="vertical-align: -5px">y=1^2+\cos(2\pi)=2.</math>
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| − | |Thus, the equation of the tangent line is  <math style="vertical-align: -5px">y=2(x-1)+2.</math>
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| − | |}
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| − | {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
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| − | !Final Answer:
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| − | :<math>\frac{dy}{dx}=2x-\sin(\pi(x^2+1))(2\pi x)</math>
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| − | :<math>y=2(x-1)+2</math>
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| − | |}
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| | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] | | [[009A_Sample_Final_1|'''<u>Return to Sample Exam</u>''']] |
