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How do you differentiate #f(x)=(x^2+x)(x^3-1)# using the product rule?

Answer 1

William Abdalla

#d f(x)=(5x^4+4x^3-2x-1)*d x#

#f(x)=(x^2+x)(x^3-1)#

#d f(x)=((x^2+x)^' * (x^3-1)+(x^3-1)^' *(x^2+x))d x#

#d f(x)=[(2x+1)(x^3-1)+3x^2(x^2+x)]* d x#

#d f(x)=[2x^4-2x+x^3-1+3x^4+3x^3]*d x#

#d f(x)=(5x^4+4x^3-2x-1)*d x#

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Answer 2

Emilia Aguilar

To differentiate ( f(x) = (x^2+x)(x^3-1) ) using the product rule, you would use the formula ( (uv)' = u'v + uv' ), where ( u = x^2 + x ) and ( v = x^3 - 1 ).

First, find ( u' ) and ( v' ):

( u' = \frac{d}{dx}(x^2 + x) = 2x + 1 )

( v' = \frac{d}{dx}(x^3 - 1) = 3x^2 )

Now, apply the product rule:

( f'(x) = (2x+1)(x^3-1) + (x^2+x)(3x^2) )

Simplify:

( f'(x) = 2x^3 - 2x + 3x^4 - 3x^2 + 3x^4 + 3x^3 )

Combine like terms:

( f'(x) = 6x^4 + 5x^3 - 5x^2 - 2x )

So, ( f'(x) = 6x^4 + 5x^3 - 5x^2 - 2x ) is the derivative of ( f(x) = (x^2+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.