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Current time:0:00Total duration:5:07

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 square root of x so we know from the definition of 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 zero and and you know some people say H approaches zero or D approach zero I just use Delta X so the change in X approaches zero and then we say f of X plus Delta X so this in this case this is f of X so it's the square root of x plus Delta X minus f of X and the switch case is square root of X all of that over the change in X over Delta X so what I'm going to do you know right now and I look at that is there's not much simplification I can do to make this come out with something meaningful I'm going to multiply this fraction times well the new I'm going to multiply the numerator and the kanji and the denominator by the conjugate of the numerator so what I mean by that let me rewrite it limit is Delta X approaches zero I'm just rewriting what I have here so I set 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 right this is just one so I can of course multiply that times if we assume that X and Delta X aren't both zero this is this is a defined number this will be one and and we can do that but this is you know 1 over 1 we're just multiplying a times this equation and we get limit as Delta X approaches 0 well this is if you view this as a minus B times a plus right let me do a little side here let me just say a plus B times a minus B is equal to a squared minus B squared right 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 was just going to be X plus Delta X right so get X plus Delta X and then what's B squared so minus square root of X is B in this you know 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's 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 in 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 one this is goes to one 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 could wait we can only do this assuming the Delta well we're dividing by Delta X to begin with so you know we know it's not 0 it's just approaching 0 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 approaches 0 we can just set Delta X is equal to 0 that's what it's approaching so that that equals 1 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 delta square root of x here plus the square root of x and that equals 1 over 2 square root of x and that equals 1/2 X to the negative 1/2 so we just proved at X to the 1/2 power the derivative of it is 1/2 X to the negative 1/2 and so it is consistent with the general the general property that the derivative of I want to know a to the or the derivative of X to the N is equal to n X to the N minus 1 even in this case where n was 1/2 well hopefully that 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 not too complicated for proof I will see you in future videos

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