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.