Difference between revisions of "009A Sample Midterm 2, Problem 5"
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Kayla Murray (talk | contribs) |
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!Step 1: | !Step 1: | ||
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− | | | + | |First, we use the Chain Rule to get |
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− | | | + | | <math>g'(x)=\cos(\cos(e^x))(\cos(e^x))'.</math> |
|} | |} | ||
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!Step 2: | !Step 2: | ||
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− | | | + | |Now, we use the Chain Rule again to get |
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+ | <math>\begin{array}{rcl} | ||
+ | \displaystyle{g'(x)} & = & \displaystyle{\cos(\cos(e^x))(\cos(e^x))'}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x)'}\\ | ||
+ | &&\\ | ||
+ | & = & \displaystyle{\cos(\cos(e^x))(-\sin(e^x))(e^x).} | ||
+ | \end{array}</math> | ||
|} | |} | ||
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| '''(a)''' <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math> | | '''(a)''' <math>3\tan^2(7x^2+5)\sec^2(7x^2+5)(14x)</math> | ||
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− | |'''(b)''' | + | | '''(b)''' <math>\cos(\cos(e^x))(-\sin(e^x))(e^x)</math> |
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|'''(c)''' | |'''(c)''' | ||
|} | |} | ||
[[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] | [[009A_Sample_Midterm_2|'''<u>Return to Sample Exam</u>''']] |