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AP.CALC:
FUN‑6 (EU)
,
FUN‑6.C (LO)
,
FUN‑6.C.1 (EK)
,
FUN‑6.C.2 (EK)

## Video transcript

let's take the derivative with respect to X of X to the n plus 1 power over n plus 1 plus some constant C and we're going to assume here because we want this expression to be defined we're going to assume that n does not equal negative 1 if it equal negative 1 we'd be dividing by 0 and that would we haven't defined what that means so let's take the derivative here so this is going to be equal to well the derivative of X to the n plus 1 over n plus 1 we could just use the power rule over here so our exponent is n plus 1 we can bring it out front so it's going to be n plus 1 times X times X to the I want to use that same color colors are the hard part times X to the instead of n plus 1 we subtract one from the exponent this is just the power rule so n plus 1 minus 1 is going to be N and then we can't forget that we have we were dividing by this n plus 1 so we have divided by n plus 1 and then we have plus C the derivative of a constant with respect to X a constant does not change as X changes so it is just going to be 0 so plus 0 and since n is not equal negative 1 we know that this is going to be defined this is just going to be something divided by itself which is just going to be 1 and this whole thing simplifies to X to the N so the derivative of this thing and this is in very general terms is equal to X to the N so given that given that what is the antiderivative let me switch colors here what is what is the antiderivative of X to the N of X to the N and remember this is just the kind of strange-looking notation we use it'll make more sense when we start doing definite integrals but what is the antiderivative of X to the N and we could say the antiderivative with respect to X if we want to and another way of calling this is the indefinite integral in definite integral well we know this is saying X to the N is the derivative of what well we just figured out it's the derivative this thing and we've written it in very general terms we're actually encapsulating multiple constants here we can have a X to the n plus 1 over n plus 1 plus 0 plus 1 plus 2 plus pi plus a billion so this is going to be equal to X to the n plus 1 over n plus 1 plus C so this is pretty powerful you can kind of view this as the reverse the reverse power rule and it applies it applies for any n as long as n does not equal negative 1 let me make that very clear n does not equal n does not equal negative 1 once again this thing would be undefined if N or equal to negative 1 so let's do a couple of examples just to apply this you could call it the reverse power rule if you want or the anti power rule so let's take the antiderivative of X to the fifth power what is the antiderivative of X to the fifth well all we have to say is a look the 5 is equal to the n we just have to increment the exponent by 1 so this is going to be equal to X to the 5 plus 1 power and then we divide by that that same value whatever the exponent was when you increment it by 1 we divide by that same value divided by 5 plus 1 and of course we want to encapsulate all of the possible anti derivatives so you put the C right over there so this is going to be equal to X to the 6th over 6 plus C and you can verify take the derivative of this using the power rule you indeed get X to the 5th let's try another one let's try and I'll do it in blue let's try the antiderivative of let's make it interesting let's make it 5 times X to the negative 2 power DX so how would we evaluate this well one simplification you can do and I'm not rigorously proven it to you just yet but we know that scalars can go in and out of the derivative operator when you're multiplying by scalars so this is indeed equal to five times the antiderivative of X to the negative 2 power DX and now we can just use you can I guess we can call it this anti power rule so this is going to be equal to five times X to the negative two power plus one plus one over the negative two power plus one plus some constant plus some constant right over here and if we and then we can rewrite this as five times negative two power plus one is X to the negative one over negative two plus one is negative one plus some constant and this is equal to this is equal to five times negative x to the negative one plus some constant and then if we want we can distribute the five so this is equal to negative five x to the negative one now we could write plus five times some constant but this is just an arbitrary constant so this is still just an arbitrary constant so maybe we could written of this if you want it to show us a different constants you could say this is C 1 C 1 C 1 you multiply 5 times C 1 you get another constant we could just call that C which is equal to 5 times C 1 but there you have a negative 5 X to the negative 1 plus C and once in all of these try to evaluate the derivative and you will see that you get you get this business right over there
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