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Current time:0:00Total duration:4:33

Video transcript

let's see if we can find the limit as H approaches zero of five log of two plus h minus five log of 2 all of that over H and I'll give you a little bit of a hint because I know you're about to pause the video and try to work through it think of your derivative properties especially the derivative of logarithmic functions especially logarithmic functions in this case with base 10 if someone just writes log without the base you can just assume that that is a 10 right over there so pause the video and see if you can work through it alright so the key here is to remember that if I have if I have f of X let me do it over here I'll do it over here f of X and I want to find f prime of let's say F prime of some number let's say a this is going to be equal to the limit as X RSR is H approaches zero is one way of thinking about it as H approaches zero of F of a plus h minus F of a all of that over H so this looks pretty close to that limit definition except we have these 5s here but lucky for us we can factor out those fives and we could factor them out we could factor them out out front here but if you just have a scalar times the expression we know from our limit properties we can actually take those out of the limit themselves so let's do that let's take both of these fives and factor them out and so this whole thing is going to simplify to 5 times the limit as H approaches 0 of log of 2 plus h minus minus log of 2 all of that over H now you might recognize what we have in yellow here let's think about it what this is if we had f of X is equal to log of X and we wanted to know what F prime of well actually let's say F prime of 2 is well this would be the limit as H approaches of log of 2 plus H 2 plus h minus log of 2 minus log of 2 all of that over H so this is really just a what we see here this by definition this right over here is is F prime of 2 if f of X is log of X this is f prime of 2 F prime of 2 so can we figure that out if f of X is log of X what is f prime of X F prime of X we don't need to use the limit definition in fact the limit definition is quite hard to evaluate this limit but we know how to take the derivative of logarithmic functions so f prime of X is going to be equal to 1 over the natural log of our base our base here we already talked about that that is 10 so 1 over natural log of 10 x times R times X if this was a natural log well then this would be 1 over natural log of e times X natural log of e is just 1 so that's where you get the 1 over X but if you have any other base you put the natural log of that base right over here in the denominator so what is f prime of 2 F prime of 2 is 1 over the natural log of 10 times 2 so this whole thing has simplified this whole thing is equal to 5 times this business so I could actually just write it as it's equal to 5 over 5 over the natural log of 10 natural log of 10 times 2 are kind of written it as 2 natural log of tens the key here for this type of exercise you might have me let me see if I can evaluate this limit be like well this looks a lot like the derivative of a logarithmic function especially the derivative when X is equal to 2 if we could just factor this these 5 out so you factor out the 5 you say this is this is the derivative of log of X when X is equal to 2 and so we know how to take the derivative of log of X if you don't know we have videos where we took where we prove this we take the derivatives of logarithms with bases other than E and you just use that to actually and you the derivative then you evaluated it too and then you're done
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