So I've been requested to do the proof of the derivative of
the square root of x, so I thought I would do a quick
video on the proof of the derivative of the
square root of x.
So we know from the definition of a derivative that the
derivative of the function square root of x, that is equal
to-- let me switch colors, just for a variety-- that's equal to
the limit as delta x approaches 0.
And you know, some people say h approaches 0,
or d approaches 0.
I just use delta x.
So the change in x over 0.
And then we say f of x plus delta x, so in this
case this is f of x.
So it's the square root of x plus delta x minus f of x,
which in this case it's square root of x.
All of that over the change in x, over delta x.
Right now when I look at that, there's not much simplification
I can do to make this come out with something meaningful.
I'm going to multiply the numerator and the denominator
by the conjugate of the numerator is what
I mean by that.
Let me rewrite it.
Limit is delta x approaching 0-- I'm just rewriting
what I have here.
So I said the square root of x plus delta x minus
square root of x.
All of that over delta x.
And I'm going to multiply that-- after switching colors--
times square root of x plus delta x plus the square root of
x, over the square root of x plus delta x plus the
square root of x.
This is just 1, so I could of course multiply that times-- if
we assume that x and delta x aren't both 0, this is a
defined number and this will be 1.
And we can do that.
This is 1/1, we're just multiplying it times this
equation, and we get limit as delta x approaches 0.
This is a minus b times a plus b.
Let me do little aside here.
Let me say a plus b times a minus b is equal to a
squared minus b squared.
So this is a plus b times a minus b.
So it's going to be equal to a squared.
So what's this quantity squared or this quantity squared,
either one, these are my a's.
Well it's just going to be x plus delta x.
So we get x plus delta x.
And then what's b squared?
So minus square root of x is b in this analogy.
So square root of x squared is just x.
And all of that over delta x times square root of x
plus delta x plus the square root of x.
Let's see what simplification we can do.
Well we have an x and then a minus x, so those
cancel out. x minus x.
And then we're left in the numerator and the denominator,
all we have is a delta x here and a delta x here, so let's
divide the numerator and the denominator by delta x.
So this goes to 1, this goes to 1.
And so this equals the limit-- I'll write smaller, because I'm
running out of space-- limit as delta x approaches 0 of 1 over.
And of course we can only do this assuming that delta--
well, we're dividing by delta x to begin with, so we know
it's not 0, it's just approaching zero.
So we get square root of x plus delta x plus
the square root of x.
And now we can just directly take the limit
as it approaches 0.
We can just set delta x as equal to 0.
That's what it's approaching.
So then that equals one over the square root of x.
Right, delta x is 0, so we can ignore that.
We could take the limit all the way to 0.
And then this is of course just a square root of x here plus
the square root of x, and that equals 1 over
2 square root of x.
And that equals 1/2x to the negative 1/2.
So we just proved that x to the 1/2 power, the derivative of it
is 1/2x to the negative 1/2, and so it is consistent with
the general property that the derivative of-- oh I don't
know-- the derivative of x to the n is equal to nx to the n
minus 1, even in this case where the n was 1/2.
Well hopefully that's satisfying.
I didn't prove it for all fractions but this is a start.
This is a common one you see, square root of x, and
it's hopefully not too complicated for proof.
I will see you in future videos.