Derivatives expressed as limits | Advanced derivatives | AP Calculus BC | Khan Academy

Let's see if we can find the limit as h approaches 0 of (5 \log(2 + h) - 5 \log(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.

All right, so the key here is to remember that if I have, if I have (f(x)), let me do it over here. I'll do it over here. (f(x)), and I want to find (f') of, let's say (f') of some number, let's say (a), this is going to be equal to the limit as (h) approaches 0 of (f(a + h) - f(a)), all of that over (h).

So this looks pretty close to that limit definition, except we have these fives here; but lucky for us, we can factor out those fives. 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 that 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 \lim_{h \to 0} \frac{\log(2 + h) - \log(2)}{h}). Now, you might recognize what we have in yellow here. Let's think about it. What this is, if we had (f(x) = \log(x)) and we wanted to know what (f'(2)) is, well this would be the limit as (h) approaches 0 of (\frac{\log(2 + h) - \log(2)}{h}).

So this is really just a, what we see here, this by definition, this right over here is (f'(2)). If (f(x) = \log(x)), this is (f'(2)). So can we figure that out? If (f(x) = \log(x)), what is (f'(x))? (f'(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'(x)) is going to be equal to (\frac{1}{\ln(b)} \cdot \frac{1}{x}), where (b) is our base. Our base here, we already talked about that, that is 10. So (\frac{1}{\ln(10)} \cdot \frac{1}{x}). If this was a natural log, well then this would be (\frac{1}{\ln(e)} \cdot \frac{1}{x}). (\ln(e)) is just 1, so that's where you get the (\frac{1}{x}). But if you have any other base, you put the (\ln(b)) right over here in the denominator.

So what is (f'(2))? (f'(2) = \frac{1}{\ln(10)} \cdot \frac{1}{2} = \frac{1}{2 \ln(10)}). So this whole thing has simplified, this whole thing is equal to (5 \times \frac{1}{2 \ln(10)}).

So I could actually just write it as it's equal to (\frac{5}{2 \ln(10)}). I could have written it as (2.5 \cdot \frac{1}{\ln(10)}). The key here for this type of exercise, you might immediately, 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 these 5s out.

So you factor out the 5, you say, hey this is the derivative of (\log(x)) when (x = 2). And so we know how to take the derivative of (\log(x)). If you don't know, we have videos where we prove this; we take the derivatives of logarithms with bases other than (e), and you just use that to actually find the derivative, then you evaluate it at 2, and then you're done.