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

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Answer 1 · 2017-03-06T20:11:58

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

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

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

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

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