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So I've been requested to do the proof of the derivative of

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the square root of x, so I thought I would do a quick

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video on the proof of the derivative of the

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square root of x.

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So we know from the definition of a derivative that the

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derivative of the function square root of x, that is equal

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to-- let me switch colors, just for a variety-- that's equal to

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the limit as delta x approaches 0.

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And you know, some people say h approaches 0,

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or d approaches 0.

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I just use delta x.

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So the change in x over 0.

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

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case this is f of x.

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So it's the square root of x plus delta x minus f of x,

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which in this case it's square root of x.

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All of that over the change in x, over delta x.

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Right now when I look at that, there's not much simplification

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I can do to make this come out with something meaningful.

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I'm going to multiply the numerator and the denominator

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by the conjugate of the numerator is what

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I mean by that.

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Let me rewrite it.

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

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what I have here.

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

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square root of x.

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All of that over delta x.

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And I'm going to multiply that-- after switching colors--

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times square root of x plus delta x plus the square root of

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x, over the square root of x plus delta x plus the

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square root of x.

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This is just 1, so I could of course multiply that times-- if

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we assume that x and delta x aren't both 0, this is a

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defined number and this will be 1.

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And we can do that.

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

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

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This is a minus b times a plus b.

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Let me do little aside here.

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

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squared minus b squared.

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So this is a plus b times a minus b.

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So it's going to be equal to a squared.

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

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either one, these are my a's.

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

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So we get x plus delta x.

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And then what's b squared?

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

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So square root of x squared is just x.

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

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plus delta x plus the square root of x.

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Let's see what simplification we can do.

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

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cancel out. x minus x.

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And then we're left in the numerator and the denominator,

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all we have is a delta x here and a delta x here, so let's

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divide the numerator and the denominator by delta x.

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So this goes to 1, this goes to 1.

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And so this equals the limit-- I'll write smaller, because I'm

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running out of space-- limit as delta x approaches 0 of 1 over.

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And of course we can only do this assuming that delta--

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well, we're dividing by delta x to begin with, so we know

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it's not 0, it's just approaching zero.

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So we get square root of x plus delta x plus

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the square root of x.

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And now we can just directly take the limit

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as it approaches 0.

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We can just set delta x as equal to 0.

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That's what it's approaching.

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So then that equals one over the square root of x.

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Right, delta x is 0, so we can ignore that.

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We could take the limit all the way to 0.

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And then this is of course just a square root of x here plus

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the square root of x, and that equals 1 over

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2 square root of x.

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And that equals 1/2x to the negative 1/2.

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So we just proved that x to the 1/2 power, the derivative of it

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is 1/2x to the negative 1/2, and so it is consistent with

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the general property that the derivative of-- oh I don't

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know-- the derivative of x to the n is equal to nx to the n

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minus 1, even in this case where the n was 1/2.

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Well hopefully that's satisfying.

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I didn't prove it for all fractions but this is a start.

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This is a common one you see, square root of x, and

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it's hopefully not too complicated for proof.

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I will see you in future videos.

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