How do you find the derivative of #2e^(-0.7x)sin(2pix)#?

1 Answer
Gerardina C.
Mar 6, 2017

#=2e^(-0.7x)(-0.7sin(2pix)+2picos(2pix))#

Explanation:

You would apply the rules:

#f(x)=g(x)*h(x)->f'(x)=g'(x)*h(x)+g(x)*h'(x)#
#f(x)=e^g(x)->f'(x)=e^g(x)*g'(x)#
#f(x)=sin(g(x))->f'(x)=cos(g(x))*g'(x)#
#f(x)=kg(x)->f'(x)=kg'(x)#

and get

#y'=2e^(-0.7x)*(-0.7)*sin(2pix)+2e^(-0.7x)*cos(2pix)*2pi#

#=2e^(-0.7x)(-0.7sin(2pix)+2picos(2pix))#

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