Proof: d/dx(sqrt(x))

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.