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let's say Y is equal to the natural log of X to the X power and what we want to do is we want to find the derivative of Y with respect to X so I encourage you to pause this video and see if you could do it so when you first try to tackle this this is a little bit daunting we know how to take the derivative of constants to some X power but how do we take a derivative of some type of a function in this case the natural log to the X power and the answer here is to use some of our logarithmic properties and then we're going to do a little bit of implicit differentiation so the first thing that we want to do and actually let me rewrite this with a little bit of space so this was the natural log of X to the to the X power so the first thing I want to get rid of this as this X as an exponential and I want to be able to apply the product rule somehow and the way we're going to do that is by taking the natural log of both sides so take the natural log of both sides and you might say well why is that helpful well if I'm taking the natural log of something to an exponent well this is the same thing actually let me write this down as a property that you may or might may or may not remember from your logarithmic properties so if I have I could write log or I'll just write natural log of if I have natural log of a to the B power this is the same thing as B times the natural log of a that's just a standard logarithmic property and so by taking the natural log of both sides this exponent can now become out front and scale the natural log function so this exponent now we can bring that out front and let's just rewrite everything so we get the natural log of y is equal to so let me put that and let me put that in parenthesis so it's the natural log of y is equal to X and that X is in blue x times the natural log times the natural log of sine x times the natural log of the natural log of X the natural log of the natural log of X so there you have it by just taking the natural log of both sides and using this logarithmic property we were able to get that now you're saying well well gee how how is this actually going to be useful for us well now we can implicitly take the derivative of both sides of this and actually let me let me scoot this over to the right a little bit just so that we can I can I have space for my derivative operator so there you go glued that over and so now let's take the derivative with respect to X of both sides so let me so I'm going to take the derivative with respect to X of the left hand side and of the right hand side and of the right hand side now on the left hand side this is going to be essentially an application of the chain rule when you learn implicit differentiation it's really just application of the chain rule it's the derivative of the outside function with respect to the inside function so the natural log of Y with respect to Y the derivative of that is just going to be 1 over Y 1 over Y times the derivative of the inside function with respect to X so dy DX dy DX that is going to be equal to well this is going to get interesting a little bit actually let's let's let me do some stuff on the on the side a little bit just so well let me just the first thing we want to do here is just apply the product rule so it's the derivative of the first expression so it's just going to be 1 times the second I guess you say function so times the natural log of the natural log of X natural log of X and then plus plus the first function just x times the derivative of the second function x times the derivative of the second function what's the derivative of the natural log of the natural log of X let's do that separately so if I have if I'm trying to take the derivative with respect to X of the natural log the natural log of the natural log of X of the natural log of X well here again I can apply the chain rule the derivative of that magenta function with respect to the inside function that is going to be 1 over the natural log of X and then times the derivative of the inside function with respect to X so times 1 over X so this is equal to 1 over X natural log of X so the derivative of this second function right over here is 1 over X natural log of X 1 over X natural log of X let's see that X and that X cancels out and so we are left with we are left with 1 over Y and I'll just I'll just write all of this in this blue color so 1 over Y times the derivative of Y with respect to X is equal to C this is just the natural log of the natural log of X the natural log of the natural log of X plus 1 over the natural log of X 1 over the natural log of X and now to solve for the derivative we can multiply both sides by Y so let's do that so we're going to multiply that side by Y and we're going to multiply this side times y and what are we going to get well on the left hand side that's why we multiplied by Y we just have the derivative of Y with respect to X derivative of Y with respect to X is equal to well Y is y is our original is all this is our original thing that we had Y was equal to the natural log let me rewrite it over here Y was equal to the natural log of X to the X power so this is essentially multiplying both sides times the natural log of X to the X power so this is going to get a little bit a little bit messy here so we could write it well we could just write it the way I wrote it just now without it without it being distributed actually let me just leave it like that so it's going to be and so we deserve our drumroll right now because this is quite involved the natural log of the natural log of X plus 1 over the natural log of X all of that times the natural log of X to the X power so that was quite involved and if someone said well what is what is the derivative of y when x is equal to e so if someone says what is this equal to when X is equal to e well we could evaluate this when X is equal to e this would be the and I know they didn't and I just made that up just now so if like the original question wasn't just what is dy/dx if they said what is dy/dx when X is equal to e if that was the original question then we could evaluate it so where we just replace all of these with ease so be an e there and e there and E there and an e there and I just picked the value e because it's easy to evaluate so the natural log of E is 1 natural log of 1 e to the 0 power is 1 so all of that just becomes 0 the natural log of e is 1 so this whole expression right over here becomes 0 plus 1 over 1 so it just becomes 1 and then the natural log of e is the natural log of e is 1 and you can have 1 to the 8th power well you could raise 1/2 to any power and you're just going to get 1 so it's 1 times 1 is equal to 1 so I just want that would be fun to try to evaluate that at a edit value that would be somewhat somewhat clean