Asked by victoria
d/dx(e^sinx2x)=
a)-cos2xe^sinx2x
b)cosxe^sin2x
c)2e^sin2x
d)2cos2xe^sin2x
e)-2cos2xe^sin2x
a)-cos2xe^sinx2x
b)cosxe^sin2x
c)2e^sin2x
d)2cos2xe^sin2x
e)-2cos2xe^sin2x
Answers
Answered by
Steve
use chain rule
d/dx(e^u) = e^u du/dx
where u = sin2x
have you any ideas now?
d/dx(e^u) = e^u du/dx
where u = sin2x
have you any ideas now?
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