 Hi there and welcome back to another example of proof with mathematical induction and this time we're going to dip back into calculus Now you've all had calculus so you may not have remembered seeing mathematical induction much in calculus But actually you can use mathematical induction and calculus to do some pretty interesting things for example Actually prove the the good old-fashioned power rule for positive integer exponents. This is your favorite Calculus derivative rule of all time, of course, and we can actually prove this for positive integer exponents of x Using mathematical induction. This seems so this is going to be a little bit of a strange approach to the proof That's kind of cool, too to get this proof going I'm going to let us assume a couple of items of calculus knowledge First of all, we're actually going to assume the product rule works. Remember the product rule is a rule for differentiating products of functions. It says that the derivative of Two functions multiplied together say f and g would be the derivative of f times g Plus f times the derivative of g now You might be thinking it's kind of weird to assume the product rule works if we're going to prove the power rule Right this doesn't isn't the true that the power rule comes first in the calculus book That's true But if you think about it and back to when you learn the product rule You didn't really need the power rule to prove the product rule So in fact you could cover the product rule first if you wanted to so we're going to assume that the product rule works In this proof and we're also going to assume that the derivative of x just by itself the derivative of x is 1 and That's safe to assume as well for a couple of reasons for one thing We could always go back to our limits for example and just prove that the limit as h goes to 0 of x plus h minus x over h is 1 and that's one way we could do it or we could think of this graphically The the function y equals x is just the Straight line that has slope one that goes to the origin and so the derivative tells us the slope But so the slope is one either way that assumption is safe to make here at the beginning And it's also actually safe to assume the product rule So let's assume both of those things are true and if you wanted to prove those then we could go back and do a Little more work now. Let's move on to the actual proof of the power. Okay, so in the base case here We want to show that our predicate Which just to remind ourselves the predicate we're working with here the P of n is that the derivative of x to the nth power is Equal to n times x to the n minus 1 that is going to be our quote-unquote P of n in this proof So what we're going to do is show that in the base case here is we're going to show that P of 1 is true We want to show that the derivative of x to the first is 1 times x to the 1 minus 1 Well, let's just deal with each side individually here on the on the Left-hand side we know based on what we're allowed to assume that this derivative is equal to 1 on the right-hand side What we have here is 1 times x to the 0 and that is also equal to 1 So those two guys are equal to each other and so the base case is established again We had to make an additional assumption in this problem right here using some basic calculus knowledge But I think that's a safe assumption to make here So the base case is established So let's move on to the inductive high in the inductive step now the inductive hypothesis The thing that we're going to assume is that for some positive integer k the power will hold so the derivative of x to the k sorry x let me erase that the derivative of x to the k power is K times x to the k minus 1 this is over what we're assuming as we move forward in the proof that I've made it up to the K-th step and now I've got to pay off the gnomes again to let me up to the next step So what are we going to try to prove here? We're going to take the proposition the predicate I should say and replace the k with a k plus 1 so we need to prove that the derivative of x to the k plus 1 is Now wherever I see a k here or here or here I'm replacing this with a k plus 1 So let me do this in a different color that would be k plus 1 parenthesis times x to the That used to be a k right here, but I'm going to replace it with a k plus 1 and There was already another minus 1 there to begin with so I have to subtract off the 1 so to simplify that expression I'm going to prove that the derivative of x to the k plus 1 is k plus 1 times x to the k so Again, you got to be very careful about what you want to assume This is our assumption and this is what we're trying to show and these two things do not mix together never keep them separate So let's move on to another slide and prove what we need to prove and again We actually have three assumptions in the game at this point. We're assuming the inductive hypothesis this Stuff in here and also the two facts about the product rule Which we haven't seen that show up yet and also the rule about the derivative of x itself so let's start and we'll begin so we're trying to prove an equation we're trying to prove the the the power rule for x to the k plus 1 so let's start with just the left-hand side of that Expression eventually I know what I need to get on the right-hand side, but I don't know that I'm there yet So what I'm going to do here is pull in some things that I know I'm going to first of all work inside the square brackets. Okay inside the square brackets. I can replace x to the k plus 1 with x to the k times x Okay, just to split off one factor of x there using my exponent rules If it helps it's just to go one exponent on there and now you know what I've got here is a product of two things So let me use the product rule to break this up now The product rule says that the the derivative of my product is going to be the derivative of the first guy So derivative x to the k times x Plus x to the k times the derivative of x and this is the good old-fashioned product rule at work here Now we're actually really close to the end here because I can use my inductive hypothesis to rephrase this Okay, I have the derivative of x to the k-th power and my inductive hypothesis. I'm going to make this in green here I'll make the equal signs in blue I guess but I'll make everything else in green the inductive hypothesis says that this Expression right here is going to be k x to the k minus 1 Okay, that's the inductive hypothesis working for you everything else up here is either going to be something It was already there like the times x or the plus x to the k or an additional assumption that we for example We said that it's okay to assume that that derivative right there is equal to 1 Now let's try to just work through all the math here on On the left term of this expression I have an x times an x to the k minus 1 if I combine those two terms I'll have the k being there but the 2x is multiplied to give me k x to the k-th power Okay, that's a all this stuff comes from there and then I have plus One times x to the k so that's just this now what you notice here Is that I have a factor of x to the k and another factor of x to the k So let's pull that factor pull that factor out and I'd have k plus 1 times x to the k Well, that's what I wanted to show right I wanted to show that the derivative of x to the k plus 1 is k plus 1 times x to the k if you go back to the previous slide That's exactly what I wanted to show and I've done it I started with the left-hand side split this up using algebra and split things up this way using the product rule Which I'm allowed to assume this is using the inductive hypothesis and the fact that I know what the derivative of x is and The rest of it is algebra and then algebra So all those steps are justified and that is the end of the proof because I have done the base case already That was pretty simple and now I've done the inductive step I assumed that the product the power rule Is true for exit for n equal to k and I've proven that it's true for n equals to k plus 1 So I have climbed the calculus stairwell and actually proven the power rule which is pretty nice Now if you want to prove the power rule for non integer powers Let's we often use things like x to the three halves and this this you'd have to have a completely different proof for that But this induction proof establishes it for positive integer powers of x And so there you have it unlocked one of the mysteries of calculus. Congratulations