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.