How do you differentiate #y=e^(-5x)cos3x#?

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Answer 1 · 2017-04-19T22:02:53

#(d(gh))/dx = g'(h) + g(h')#

Let #g = e^(-5x) and h = cos(3x)#, then:

#g' = -5e^(-5x) and h' = -3sin(3x)#

Substituting into the product rule:

#(d(e^(-5x)cos(3x)))/dx = -5e^(-5x)cos(3x) -3e^(-5x)sin(3x)#

#(d(e^(-5x)cos(3x)))/dx = -e^(-5x)(5cos(3x) +3sin(3x))#