Name: Proof: d/dx(sqrt(x)) | Taking derivatives | Differential Calculus | Khan Academy
Uploaded: 2022-07-12T12:47:58.000Z
Duration: 5 min 7 s
Description: Thousands of YouTube videos with English-Chinese subtitles! Now you can learn to understand native speakers, expand your vocabulary, and improve your pronunciation...

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

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

And you know, some people say h approaches 0,

And then we say f of x plus delta x, so in this

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

Limit is delta x approaching 0-- I'm just rewriting

So I said the square root of x plus delta x minus

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

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

This is 1/1, we're just multiplying it times this

equation, and we get limit as delta x approaches 0.

Let me say a plus b times a minus b is equal to a

So what's this quantity squared or this quantity squared,

Well it's just going to be x plus delta x.

So minus square root of x is b in this analogy.

And all of that over delta x times square root of x

Well we have an x and then a minus x, so those

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.

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

So we get square root of x plus delta x plus

And now we can just directly take the limit

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

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.

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.