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

Answer 1

#f'(x)=2(2x^2+1)+4x(2x-1)#

#=12x^2-4x+2#

Product rule states that #d/dx(f(x)*g(x))=f(x)*g'(x)+g(x)*f'(x)#.

Applying it to this specific situation results in

#f'(x)=2(2x^2+1)+4x(2x-1)#
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Answer 2

To differentiate ( f(x) = (2x^2 + 1)(2x - 1) ) using the product rule:

Let ( u(x) = 2x^2 + 1 ) and ( v(x) = 2x - 1 ).

Apply the product rule, which states that if ( f(x) = u(x)v(x) ), then ( f'(x) = u'(x)v(x) + u(x)v'(x) ).

[ u'(x) = \frac{d}{dx}(2x^2 + 1) = 4x ] [ v'(x) = \frac{d}{dx}(2x - 1) = 2 ]

Using the product rule:

[ f'(x) = (4x)(2x - 1) + (2x^2 + 1)(2) ] [ f'(x) = 8x^2 - 4x + 4x^2 + 2 ] [ f'(x) = 12x^2 - 4x + 2 ]

So, the derivative of ( f(x) = (2x^2 + 1)(2x - 1) ) is ( f'(x) = 12x^2 - 4x + 2 ).

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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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