How do you differentiate #f(x)=(x-e^x)(cosx+2x)# using the product rule?

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Answer 1 · 2016-05-27T08:17:29

#=ex\sin (x)-4ex-x\sin (x)+4x+\cos (x)-e\cos(x)#

#\frac{d}{dx}((x-ex)(cos (x)+2x))#

Applying product rule,
#(f\cdot g)^'=f^'\cdot g+f\cdot g^'#

#f=x-ex,\g=\cos (x)+2x#

#=\frac{d}{dx}(x-ex)(\cos (x)+2x)+\frac{d}{dx}(cos (x)+2x)(x-ex\)#

We know,
#\frac{d}{dx}(x-ex)=1-e#
#\frac{d}{dx}(\cos \(x)+2x)=-\sin \(x)+2#

Finally,
#=(1-e)\(\cos (x\)+2x\)+\(sin (x)+2\)(x-ex)#

Simplifying it,
#=ex\sin (x)-4ex-x\sin (x)+4x+\cos (x)-e\cos(x)#