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Current time:0:00Total duration:9:02

Video transcript

a bit of a classic implicit differentiation problem is the problem why I don't want to write it that thick y is equal to X to the X and then to find out what the derivative of Y is with respect to X and people look at that oh I have X to it you know I don't have just a constant exponent here so I can't just use the power rules how do you do it and the trick here is really just to take the natural log of both sides of this equation and this is going to build up to what we're going to do later in this video so if you take the natural log of both sides of this equation you get you get the natural log of y is equal to the natural log of X to the X now our power rules or I guess our natural log rules say look if I'm taking the natural log of something to the something this is equivalent to I can rewrite the natural log of X to the X as being equal to x times the natural log of X so let me rewrite everything again so this if I take the natural log of both sides of that equation I get the natural log of Y is equal to x times the natural log of X and now we can take the derivative of both sides of this with respect to X so the derivative with respect to X of that and then the derivative with respect to X of that now this is we're going to apply a little bit of our a little bit of the chain rule so the chain rule what's the derivative of this with respect to X what's the derivative of our inner expression with respect to X is a little implicit differentiation so it's d y with respect to x times the derivative of the whole thing with respect to this inner function so the derivative of natural log of X is 1 over X the derivative of natural log of Y with respect to Y is 1 over Y so times 1 over Y and then that is equal to the derivative of this this is just the product rule and I'll arbitrarily switch colors here it's the derivative of the first term which is 1 times the second term so times the natural log of plus the derivative of the second term which is 1 over x times the first term so times X and so we get we get dy/dx times 1 over y is equal to natural log of X plus this is just turns out to be 1 right X divided by X and then you multiply both sides of this by why you get dy/dx is equal to Y times the natural log of X plus 1 and if you don't like this y sitting here you can just make this substitution Y is equal to X to the X so you can say that the derivative of Y with respect to X is equal to X to the x times the natural log of X plus 1 and that's a fun problem and this is often kind of given as a a trick problem or sometimes even a bonus problem if people don't know to take the natural log of both sides of that but I was given an even more difficult problem and that's what we're going to tackle in this but it's good to see this problem done first because it gives us the basic tools so the more difficult problem we're going to deal with is this one let me write it down so the problem is y is equal to X to the and here's the twist X to the X to the X and we want to find out dy/dx we want to find out the derivative of Y with respect to X so to solve this problem we essentially use the same tool we use the natural log to essentially break down this exponent and get into the terms that we can deal with so we can use the product rule so let's take the natural log of both sides of this equation like we did last time you get the natural log of Y is equal to the natural log of X to the X to the X X to the X to the X and this is just the exponent on this so we can rewrite this as X to the x times the natural log times the natural log of X now well immediately so now our sitar expression or equation has simplified to the natural log of y is equal to X to the x times the natural log of X but we still have this nasty X to the X here we we know no easy way to take the derivative there although I've actually just shown you what the derivative of this is so we could actually just apply it right now I read I was going to take the natural log again and turn to this big messy confusing thing but I realize that earlier in this video I just solved for what the derivative of X to the X is it's this thing right here it's this crazy expression right here so we just have to we just have to remember that and then apply and then do our problem so let's do our problem and if we hadn't solved this ahead of time if we hadn't solved this ahead of time it was kind of a unexpected benefit of doing the simpler version of the problem you could just take keep taking the natural log of this and just but it will just get a little bit Messier but since we already know what the derivative of X to the X is let's just apply it so if we take the derivative we take the derivative of both sides equation derivative of this is equal to the derivative of this we'll ignore this for now derivative of this with respect to X is the derivative of the natural log of Y with respect to Y so that's 1 over Y times the derivative of Y with respect to X that's just the chain rule we learn about an implicit differentiation and so this is equal to this is equal to the derivative of the first term times the second term and I'm going to write it out here just because I don't want to skip steps and confuse people so this is equal to the derivative with respect to X of x to the x times the natural log of X plus the derivative with respect to X of the natural log of x times X to the X X to the X so let's see let's do that let's focus on the right hand side of this equation what is the derivative of X to the X with respect to X well we just solve that problem right here it's X to the X natural log of X plus 1 so this piece right there I'll do it in a different color that piece right there I already forgot what it was it was X to the X natural log of X plus 1 that is X to the x times the natural log of X plus 1 and then we're going to multiply that times the natural log of x times the natural log of X and then we're going to add that 2 plus the derivative of the natural log of X so plus the derivative of natural log of X that's fairly straightforward that's 1 over X times X to the X times X to the X and of course the left-hand side of the equation was just 1 over Y dy DX and we can multiply both sides of this now by Y and we get dy DX is equal to Y times all of this crazy stuff X to the x times the natural log of X plus 1 times the natural log of X plus 1 over X tyonne times X to the X that's X to the negative 1 we could rewrite this as X to the minus 1 and then you add the exponents you could write this as X to the X minus 1 power and if we don't like this Y here if we don't like this Y we can just substitute it back Y was equal to this this crazy thing right there so our final answer for this seemingly well on one level looks like a very simple problem but on another level when you appreciate what there what it's saying it's like oh this is a very complicated problem you get the derivative of Y with respect to X is equal to Y which is this will just back substitute so that's X to the X to the x times all of this stuff times and I'll just put it in green times X to the X natural log of X plus 1 times the natural log of X and then all of that plus X to the X minus 1 so who would have thought sometimes math is elegant you take the derivative of something like this and you get something neat for example when you take the derivative of natural log you get 1 over natural log of X you get 1 over X that's very simple and elegant and it's nice that math worked out that way but sometimes you do something you take an operation on something that looks pretty simple and elegant and you get something that's hairy and and not that pleasant to look at but it's a pretty interesting problem there you go