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# Démonstration de la formule de dérivation de x puissance n

Transcription de la vidéo

I just did several videos on
the binomial theorem, so I think, now that they're done, I
think now is good time to do the proof of the derivative
of the general form. Let's take the derivative
of x to the n. Now that we know the
binomial theorem, we have the tools to do it. How do we take the derivative? Well, what's the classic
definition of the derivative? It is the limit as delta x
approaches zero of f of x plus delta x, right? So f of x plus delta x in this
situation is x plus delta x to the nth power, right? Minus f of x, well f of x
here is just x to the n. All of that over delta x. Now that we know the binomial
theorem we can figure out what the expansion of x plus
delta x is to the nth power. And if you don't know the
binomial theorem, go to my pre-calculus play list
and watch the videos on the binomial theorem. The binomial theorem tells us
that this is equal to-- I'm going to need some space for
this one-- the limit as delta x approaches zero. And what's the
binomial theorem? This is going to be equal to--
I'm just going to do the numerator-- x to the
n plus n choose 1. Once again, review the binomial
theorem if this is looks like latin to you and you
don't know latin. n choose 1 of x to the n minus
1 delta x plus n choose 2 x to the n minus 2, that's x n
minus 2, delta x squared. Then plus, and we have a bunch
of the digits, and in this proof we don't have to go
through all the digits but the binomial theorem tells us what
they are and, of course, the last digit we just keep adding
is going to be 1-- it would be n choose n which is 1. Let me just write that
down. n choose n. It's going to be x to the
zero times delta x to the n. So that's the
binomial expansion. Let me switch back to minus,
green that's x plus delta x to the n, so minus
x to the n power. That's x to the n, I know
I squashed it there. All of that over delta x. Let's see if we can simplify. First of all we have an x to
the n here, and at the very end we subtract out an x to the
n, so these two cancel out. If we look at every term here,
every term in the numerator has a delta x, so we can divide
the numerator and the denominator by delta x. This is the same thing
as 1 over delta x times this whole thing. So that is equal to the limit
as delta x approaches zero of, so we divide the top and the
bottom by delta x, or we multiply the numerator
times 1 over delta x. We get n choose 1 x
to the n minus 1. What's delta x divided by
delta x, that's just 1. Plus n choose 2, x
to the n minus 2. This is delta x squared, but
we divide by delta x we just get a delta x here. Delta x. And then we keep having a bunch
of terms, we're going to divide all of them by delta x. And then the last term is
delta x to the n, but then we're going to divide
that by delta x. So the last term becomes n
choose n, x to the zero is 1, we can ignore that. delta x
to the n divided by delta x. Well that's delta x
to the n minus 1. Then what are we doing now? Remember, we're taking
the limit as delta x approaches zero. As delta x approaches zero,
pretty much every term that has a delta x in it,
it becomes zero. When you multiply but
zero, you get zero. This first term has no
delta x in it, but every other term does. Every other term, even after
we divided by delta x has a delta x in it. So that's a zero. Every term is zero, all
of the other n minus 1 terms, they're all zeros. All we're left with is that
this is equal to n choose 1 of x the n minus 1. And what's n choose 1? That equals n factorial over 1
factorial divided by n minus 1 factorial times x
to the n minus 1. 1 factorial is 1. If I have 7 factorial divided
by 6 factorial, that's just 7. Or if I have 3 factorial
divided by 2 factorial, that's just 3, you can work it out. 10 factorial divided by
9 factorial that's 10. So n factorial divided
by n minus 1 factorial, that's just equal to n. So this is equal to n
times x to the n minus 1. That's the derivative
of x to the n. n times x to the n minus 1. We just proved the derivative
for any positive integer when x to the power n, where n
is any positive integer. And we see later it actually
works for all real numbers and the exponent. I will see you in
a future video. Another thing I wanted to point
out is, you know I said that we had to know the
binomial theorem. But if you think about it, we
really didn't even have to know the binomial theorem because we
knew in any binomial expansion-- I mean, you'd have
to know a little bit-- but if you did a little
experimentation you would realize that whenever you
expand a plus b to the nth power, first term is going to
be a to the n, and the second term is going to be plus n a to
the n minus 1 b. And then you are going to keep
having a bunch of terms. But these are the only terms
that are relevant to this proof because all the other terms get
canceled out when delta x approaches zero. So if you just knew that you
could have done this, but it's much better to do it with
the binomial theorem. Ignore what I just said
if it confused you. I'm just saying that we could
have just said the rest of these terms all go to zero. Anyway, hopefully you
found that fulfilling. I will see you in
future videos.