How do you differentiate #f(x)=4xe^xsinx# using the product rule?

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Answer 1 · 2016-03-19T09:33:21

#d/(dx)f(x)=4*e^x(sinx+xsinx+xcosx)#

#d/(dx)(f(x)g(x)h(x))#
= #(df(x))/(dx)*g(x)h(x)+(dg(x))/(dx)*f(x)h(x)+(dh(x))/(dx)*f(x)g(x)#

Hence, #d/(dx)(4x*e^x*sinx)#

= #(d(4x))/(dx)*e^x*sinx+(de^x)/(dx)*4x*sinx+(dsinx)/(dx)*4x*e^x#

= #4*e^x*sinx+e^x*4x*sinx+cosx*4x*e^x#

= #4*e^x(sinx+xsinx+xcosx)#

How do you differentiate #f(x)=4x*e^x*sinx# using the product rule?