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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.