How do you differentiate #g(x) =e^xsinx# using the product rule?

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Answer 1 · 2016-06-10T19:16:44

#g'(x) = e^xsinx+e^xcosx#

The product rule tells us that #d/dx(color(red)(u)*color(blue)(v)) = color(red)((du)/dx)*color(blue)(v)+color(red)(u)color(blue)((dv)/dx)#

In this question,
#color(red)u = color(red)(e^x)#, so #color(red)((du)/dx) = color(red)(e^x)#,

and

#color(blue)(v) = color(blue)(sinx)#, so #color(blue)((dv)/dx) = color(blue)(cosx)#.

#g'(x) = color(red)(e^x)*color(blue)(sinx)+color(red)(e^x)color(blue)(cosx)#

#= e^xsinx+e^xcosx#

Factor the #e^x# if desired.