# How do you differentiate #y=(e^x(x^2-7))#?

You have to use the Product Rule

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To differentiate ( y = (e^x)(x^2 - 7) ), you can use the product rule of differentiation.

The product rule states that if you have two functions ( u(x) ) and ( v(x) ), the derivative of their product is given by:

[ \frac{d}{dx}(u(x)v(x)) = u'(x)v(x) + u(x)v'(x) ]

Applying the product rule to ( y = (e^x)(x^2 - 7) ), where ( u(x) = e^x ) and ( v(x) = x^2 - 7 ):

[ y' = (e^x)(x^2 - 7)' + e^x(x^2 - 7)' ]

Now, differentiate each term separately:

- ( (x^2 - 7)' = 2x )
- ( (e^x)' = e^x )

So,

[ y' = (e^x)(2x) + e^x(2x) ] [ y' = 2xe^x + 2xe^x ] [ y' = 4xe^x ]

Therefore, the derivative of ( y = (e^x)(x^2 - 7) ) is ( y' = 4xe^x ).

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To differentiate ( y = e^x(x^2 - 7) ), you would use the product rule of differentiation.

First, differentiate ( e^x ), which is simply ( e^x ). Then, apply the product rule to differentiate ( x^2 - 7 ). The derivative of ( x^2 - 7 ) is ( 2x ).

So, the derivative of ( y ) with respect to ( x ) is:

[ \frac{dy}{dx} = e^x(x^2 - 7) + e^x(2x) ]

[ = e^x(x^2 - 7 + 2x) ]

[ = e^x(x^2 + 2x - 7) ]

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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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