How do you differentiate #f(x)=e^(3x)cos2x# using the product rule?

1 Answer

Answer:

#f' (x)=e^(3x)*[3*cos 2x-2*sin 2x]#

Explanation:

Given #f(x)=e^(3x)*cos(2x)#

Use the formula
#d/dx(uv)=u*d/dx(v)+v*d/dx(u)#
Let #u=e^(3x)# and #v=cos(2x)#

#d/dx(uv)=u*d/dx(v)+v*d/dx(u)#
#f' (x)=d/dx(e^(3x)*cos(2x))=e^(3x)*d/dx(cos(2x))+cos(2x)*d/dx(e^(3x))#

#f' (x)=e^(3x)*-sin(2x)*d/dx(2x)+cos(2x)*e^(3x)*d/dx(3x)#

#f' (x)=e^(3x)*(-sin(2x))(2)+cos(2x)e^(3x)(3)#

simplify by factoring common factors

#f' (x)=e^(3x)*[3*cos 2x-2*sin 2x]#

God bless...I hope the explanation is useful.