How do you differentiate #f(x)=x^3(4x - 12)# using the product rule?

Answer 1
Product Rule: If we have two functions #f(x)# and #g(x)#, then #d/dx# #f(x)g(x)# = #f^'(x)g(x) + f(x)g^'(x)#
Original Equation: #f(x)=x^3(4x-12)#
The two equations you're multiplying are #x^3# and #(4x-12)#
Take the derivative of the first equation (#x^3#) and times it by the second equation (#(4x-12)#)
You should get #(3x^2)•(4x-12)#
Next, take the derivative of the second equation (#(4x-12)#) and times it by the first (#x^3#).
You should get #(x^3)•(4)#

Add them all together now.

You should get #((3x^2)•(4x-12) )+ ((x^3)•(4))#

Add like terms and multiply it out.

Your final answer should be #16x^3-36x^2#
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Answer 2

To differentiate f(x)=x^3(4x - 12) using the product rule, follow these steps:

  1. Identify the two functions being multiplied: u(x) = x^3 and v(x) = (4x - 12).
  2. Apply the product rule: (f'(x) = u'(x)v(x) + u(x)v'(x)).
  3. Calculate the derivatives of u(x) and v(x): (u'(x) = 3x^2) and (v'(x) = 4).
  4. Substitute the derivatives and original functions into the product rule formula.
  5. Simplify the expression to get the final result for (f'(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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