How do you differentiate #(2x^2-4x+4)e^x#?

Answer 1
for a product of functions you have the formula #if f(x)=g(x)*h(x)# #if f'(x)=g'(x)*h(x)+g(x)*h'(x)#
so lets start :) your functions are #g(x)=2x^2-4x+4# and #h(x)=e^x#
so #f'(x)=(4x-4)*e^x+(2x^2-4x+4)*e^x=(4x-4+2x^2-4x+4)*e^x=2x^2e^x#
remember that #d/dxe^u=e^u*d/dxu# in this case as u=x; du/dx=1
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Answer 2

To differentiate the expression (2x^2 - 4x + 4)e^x, you can use the product rule of differentiation.

The product rule states that if you have two functions, u(x) and v(x), their derivative with respect to x, denoted as u'(x) and v'(x) respectively, then the derivative of the product u(x)v(x) is given by:

(uv)' = u'v + uv'

Applying the product rule to the given expression:

u(x) = (2x^2 - 4x + 4) v(x) = e^x

Differentiating u(x) and v(x):

u'(x) = d/dx(2x^2 - 4x + 4) v'(x) = d/dx(e^x)

Now, find the derivatives:

u'(x) = 4x - 4 v'(x) = e^x

Then, apply the product rule:

(2x^2 - 4x + 4)e^x = (u(x)v(x))' = u'(x)v(x) + u(x)v'(x)

= (4x - 4)e^x + (2x^2 - 4x + 4)e^x

= 4xe^x - 4e^x + 2x^2e^x - 4xe^x + 4*e^x

= (2x^2 - 4)e^x + 4xe^x - 4e^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.

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.

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.

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.

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