How do you differentiate #g(y) =e^x(x^3-1) # using the product rule?

Answer 1

#g'(y)=0#

#g'(x)=3x^2e^x+(x^3-1)e^x#

According to the product rule,

#d/dx[f(x)*g(x)]=f(x)*g'(x)+g(x)*f'(x)#.

When this rule is applied to this specific function, it produces

Case 1: g(y) as it appears in the query

The function only contains x terms and so the derivative #g'(y)=0#.

Case 2: The product rule is then applicable if g(x) is assumed.

#g'(x)=3x^2e^x+(x^3-1)e^x#
#=e^x(3x^2+x^3-1)#
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Answer 2

To differentiate ( g(y) = e^x(x^3 - 1) ) using the product rule, you apply the formula:

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

Where ( u = e^x ) and ( v = x^3 - 1 ).

( u' ) represents the derivative of ( u ) with respect to ( x ), and ( v' ) represents the derivative of ( v ) with respect to ( x ).

The derivative of ( e^x ) with respect to ( x ) is ( e^x ).

The derivative of ( x^3 - 1 ) with respect to ( x ) is ( 3x^2 ).

Now, applying the product rule:

[ g'(x) = e^x(3x^2) + e^x(x^3 - 1) ]

[ g'(x) = 3x^2e^x + (x^3 - 1)e^x ]

This is the result of differentiating ( g(y) = e^x(x^3 - 1) ) using the product rule.

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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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