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Avihu R.

asked • 01/08/15

how do i find the derivative of 2xe^x^2?

I have an equation based on the principle
  lim     f(x)            lim     f'(x)
x→x0  g(x)    =    x→x0  g'(x)
 
 lim     sin2x   =    lim     2sinxcosx   =        cos2x lx=0      =   1
x→0    ex²-1        x→0       2xex²             2ex²(2x2+1) lx=0      6
 
I don't understand how the transition is made from the second step to the third.
Do I say that f(x)=2x and g(x)=e^x2? how do i get g'(x)??
Thanks!

2 Answers By Expert Tutors

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Michael F. answered • 01/10/15

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Donald S. answered • 01/08/15

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