Hey, everyone. We just learned a rule for finding the derivative of our general exponential function, \( b^x \). Here, we're going to take a look at finding the derivative of our general logarithmic function. That's \( \log_b(x) \). Now there is, of course, a rule that we can use in order to find this derivative.
Now here, I'm going to walk you through sort of where this rule comes from using what we already know about exponential functions to give you a little bit of intuition as to why this rule is what it is. Once we actually know this rule, we can just apply it to some examples, and we'll work through those together. So let's go ahead and jump into things. Now looking at this equation, \( y = \log_b(x) \), my ultimate goal here is to determine my derivative, \( \frac{dy}{dx} \). We want to find what that is.
Now this equation is currently written in its logarithmic form, but I can rewrite this in its equivalent exponential form using properties of logs and exponents, taking this base and raising it to the power of \( y \). This is then equal to \( x \). So I can rewrite this as \( x = b^y \). Now since this is an exponential function, we can then use what we already know about finding the derivative of exponential functions. So we can differentiate both sides of this equation in order to ultimately find the derivative of our logarithmic function that we're looking for.
That's \( \frac{dy}{dx} \). So having rewritten \( y = \log_b(x) \) as \( x = b^y \), we now want to take the derivative on both sides here. Now if you're familiar with the process of implicit differentiation, that's exactly what we're doing here. But if you aren't familiar with that process, all we're doing is just taking the derivative on either side using all the rules that we already know, so don't worry about it. Now looking at this, the derivative of \( x \) with respect to \( x \), we know that this is just equal to 1.
Then coming over to our right side, the derivative of \( b^y \), I know that I can use my rule for finding the derivative of an exponential function here. But since this power is not \( x \), it is \( y \), that means I need to apply the chain rule when I'm taking this derivative. So I have \( b^y \times \ln(b) \) using my rule here. And then I want to multiply this by the derivative of that inside function \( y \) since I know that that is a function of \( x \). This is \( \frac{dy}{dx} \).
Now this is ultimately the derivative that we want to find. We want to find the derivative of our logarithmic function. That's \( \frac{dy}{dx} \). So we want to solve for this \( \frac{dy}{dx} \), which we can do by isolating it, dividing this over, this \( b^y \times \ln(b) \), in order to get it to cancel. Then we're just left with that \( \frac{dy}{dx} \).
Now swapping the sides of my equation here, this gives me that \( \frac{dy}{dx} = \frac{1}{b^y \times \ln(b)} \). And this is the exact derivative that we were looking for. Now, typically, when we're working with derivatives with respect to \( x \), we don't usually want that derivative in terms of \( y \). But we know exactly what \( y \) is. It's \( \log_b(x) \).
So we can just plug that right in here. So this is \( b^{\log_b(x)} \times \ln(b) \), just having replaced that \( y \) with this \( \log_b(x) \). But using properties of logs and exponents, if I take \( b \) and raise it to the power of \( \log_b \), that just cancels out. So all I'm left with here is just \( x \).
So then \( \frac{dy}{dx} \), my derivative of \( \log_b(x) \), is equal to \( \frac{1}{x \times \ln(b)} \). That's what's left over in my denominator here. And this is my full rule for finding the derivative of a general logarithmic function. The derivative of \( \log_b(x) \) is \( \frac{1}{x \times \ln(b)} \). Now these restrictions on \( b \) are the same exact ones that we saw when dealing with our exponential function.
And here, \( x \) must also be greater than 0 because we know we can't take the log of a negative number, and it does end up in the denominator of my fraction. So, hopefully, this gave you a little bit of context as to why this rule is what it is. But now that we have this rule, we can just apply it to some examples and actually put it to use. So let's work through our first example here and find the derivative of this function, \( g(x) = \log_8(x) \). So applying my rule here in order to find \( g'(x) \), my derivative, this is then going to be equal to \( \frac{1}{x \times \ln(8)} \).
And this is my full derivative. \( g'(x) = \frac{1}{x \times \ln(8)} \). Now we're going to have to deal with finding the derivative of some more complicated functions involving these logarithmic functions. If at this point you have not yet learned the chain rule, you should stop here and just focus on applying this rule to some basic logs. But for the rest of you, let's continue on in finding the derivative of this next function, \( f(x) \).
This is \( \log_5(x^2) \). So here, we can see that we're not just taking \( \log_5(x) \). We're taking \( \log_5(x^2) \). Because of that, we need to use the chain rule because we have a function inside of our logarithmic function. So in order to find this derivative, \( f'(x) \), we're going to use the rule that we just learned but along with the chain rule.
So when starting with this outside function, the derivative of a \( \log_5(x^2) \), when we take this derivative, we cannot just put an \( x \) there. We have to put the full function that we started with. So this is then \( \frac{1}{x^2} \), not just \( x \), times the natural log of our base, which is just 5. Then from here, applying the chain rule, we multiply this by the derivative of that inside function. Our inside function is, in this case, \( x^2 \).
So the derivative of \( x^2 \) is just 2x. So we multiply this by 2x. That allows these \( x \)'s to cancel, and I am left here with \( \frac{2}{x \times \ln(5)} \) for my final derivative here, \( f'(x) \), having applied both the chain rule and our new rule for finding the derivative of logarithmic functions. Now we're going to continue getting practice putting all of our rules for derivatives all together. I'll see you in the next one.