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How do you differentiate #f(x)=(2x^2-4x+4)e^x# using the product rule?

Answer 1

Emily Ammerman

#f'(x)=2x^2e^x#

The product rule states that #d/dx[f(x)*g(x)]=f(x)*g'(x)+g(x)*f'(x)#.

When combined with other accepted differentiation rules, this rule produces

#d/dx(2x^2-4x+4)e^x=(2x^2-4x+4)e^x+(4x-4)e^x#

#=e^x(2x^2)#

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Answer 2

Axel Alvarez

To differentiate ( f(x) = (2x^2 - 4x + 4)e^x ) using the product rule, follow these steps:

Identify the functions ( u(x) ) and ( v(x) ): ( u(x) = 2x^2 - 4x + 4 ) ( v(x) = e^x )
Differentiate ( u(x) ) and ( v(x) ) separately: ( u'(x) = 4x - 4 ) ( v'(x) = e^x )
Apply the product rule: ( f'(x) = u'(x)v(x) + u(x)v'(x) ) ( f'(x) = (4x - 4)e^x + (2x^2 - 4x + 4)e^x )
Simplify the expression: ( f'(x) = (4x - 4 + 2x^2 - 4x + 4)e^x ) ( f'(x) = (2x^2 - 4)e^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.