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# Find the derivative of the function using the definition of derivative. State the domain of the function and the domain of its derivative.$f(x) = x^2 - 2x^3$

## $2 x-6 x^{2}$

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### Video Transcript

in this problem you were trying to determine the derivative of the function X squared minus two X cubed. And you want to do so using the definition of a derivative and the definition of the derivative is a limit definition and it would be f prime of X equals The limit as delta x approaches zero of the function evaluated at X plus delta X minus the original function all over delta X. So that means we have to first evaluate the function at X plus delta X. So it would read F prime of X equals the limit As Delta X approaching zero of X plus delta X quantity squared minus two times X plus delta X quantity cute. So that took care of this part and now we're going to subtract the original function and the original function is X squared minus two X cubed and that is all over delta X. So we now need to square the spinal meal and cube this spinal meal. So we will get F prime of X equals the limit As Delta X is approaching zero of x squared plus two X delta X plus delta X squared -2. Now we're going to cube that. I know yeah I know meal and we're gonna get x cubed plus three X squared delta X plus three X delta X squared plus delta X cubed. And we're gonna subtract X squared minus two X. Cute. And that's all over delta X. We now want to distribute this negative too. So we will get F prime of X Is equal to the limit as Delta X approaches zero of x squared plus two X delta X plus delta X squared minus two X cubed minus six X squared delta x minus six X delta x squared minus two delta x cubed. We're also going to distribute this negative. So we'll have minus X squared plus two X cubed all over delta X. We have some like terms that can now cancel X squared and negative X squared will cancel. Two X cubed and negative. Two X cube will cancel. So we now have F prime is equal to the limit as delta X is approaching zero. Uh two X delta X plus delta X squared minus six X squared delta X minus six X delta X squared minus two delta X cubed. All over delta X. And if you look carefully at that numerator, you will see that they have a greatest common factor of a delta X. So we could now factor out that greatest common factor of delta X. And you would be left with two X plus delta x minus six X squared minus six X delta x minus two delta X squared. And that's all over delta X. Which now the delta X. That you factored out will reduce and cancel with that denominator leaving you with f prime of X equals the limit. As Delta X is approaching zero of two X plus delta x minus six X squared minus six X. Delta x minus two delta x squared. We can now do direct substitution and we could let the delta X approach the zero. So in essence we're substituting zero in for all of those delta excess. And you will arrive at you were derivative. So we will have two X plus zero minus six X squared minus six X times zero minus two times zero squared for a final derivative of two x minus six X squared.

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