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

Answer 1

Christopher Allers

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

Product rule stipulates that

#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)#

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Answer 2

Charles Anctil

To differentiate ( f(x) = 4x \cdot e^x \cdot \sin(x) ) using the product rule:

Identify the functions being multiplied: ( u(x) = 4x ) and ( v(x) = e^x \cdot \sin(x) ).
Apply the product rule: ( f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x) ).
Find the derivatives of ( u(x) ) and ( v(x) ): ( u'(x) = 4 ) and ( v'(x) = e^x \cdot \sin(x) + e^x \cdot \cos(x) ).
Substitute into the product rule formula: ( f'(x) = 4 \cdot (e^x \cdot \sin(x)) + (4x) \cdot (e^x \cdot \sin(x) + e^x \cdot \cos(x)) ).
Simplify the expression: ( f'(x) = 4e^x \cdot \sin(x) + 4x \cdot e^x \cdot \sin(x) + 4x \cdot e^x \cdot \cos(x) ).

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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.

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

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