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MIT 18.01 Single Variable Calculus (OCW)
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OpenStudy (anonymous):
The problem for session 4: How can I calculate f'(x) of f(x)=sin2x as 2cos2x? Have I missed anything?
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OpenStudy (stacey):
This uses the chain rule. the derivative of sin x is cos x, but we do have sin x. We have sin 2x.
OpenStudy (stacey):
Taking the derivative will give us cos 2x times the derivative of 2x. Since the derivative of 2x is 2, the answer is (cos 2x)*2, or 2*cos 2x
OpenStudy (stacey):
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