Home > Calculus > Product Rule

How do you differentiate #f(x)=xe^(2x)sinx# using the product rule?

Answer 1

Samantha Adkison

#f'(x)=e^{2x}(2x\sin x+x\cos x+\sin x)#

Given function:

#f(x)=xe^{2x}\sin x#

differentiating above function using product rule as follows

#d/dxf(x)=d/dx(xe^{2x}\sin x)#

#f'(x)=xe^{2x}d/dx(\sin x)+x\sin xd/dx(e^{2x})+e^{2x}\sin xd/dx(x)#

#=xe^{2x}\cos x+x\sin x(2e^{2x})+e^{2x}\sin x(1)#

#=xe^{2x}\cos x+2xe^{2x}\sin x+e^{2x}\sin x#

#=e^{2x}(2x\sin x+x\cos x+\sin x)#

By signing up, you agree to our Terms of Service and Privacy Policy

Answer 2

Ezekiel Alexander

To differentiate ( f(x) = xe^{2x}\sin(x) ) using the product rule, we apply the rule which states that the derivative of the product of two functions ( u(x) ) and ( v(x) ) is given by ( u'(x)v(x) + u(x)v'(x) ). Here, ( u(x) = x e^{2x} ) and ( v(x) = \sin(x) ).

Applying the product rule:

( f'(x) = u'(x)v(x) + u(x)v'(x) )

( f'(x) = (1 \cdot e^{2x} + x \cdot (2e^{2x}))\sin(x) + x e^{2x} \cdot \cos(x) )

( f'(x) = e^{2x}\sin(x) + 2xe^{2x}\sin(x) + x e^{2x} \cdot \cos(x) )

By signing up, you agree to our Terms of Service and Privacy Policy

Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.