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December 18, 2014

December 18, 2014

Posted by **victoria** on Monday, February 20, 2012 at 9:50am.

a)-cos2xe^sinx2x

b)cosxe^sin2x

c)2e^sin2x

d)2cos2xe^sin2x

e)-2cos2xe^sin2x

- calculus -
**Steve**, Monday, February 20, 2012 at 11:54amuse chain rule

d/dx(e^u) = e^u du/dx

where u = sin2x

have you any ideas now?

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