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

d/dx(e^sinx2x)=

a)-cos2xe^sinx2x
b)cosxe^sin2x
c)2e^sin2x
d)2cos2xe^sin2x
e)-2cos2xe^sin2x

  • calculus - , Monday, February 20, 2012 at 11:54am

    use chain rule

    d/dx(e^u) = e^u du/dx
    where u = sin2x

    have you any ideas now?

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