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# Proof of power rule for square root function

## Video transcript

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.

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