 So now that we've learned about the power rule, and we've learned about logarithmic differentiation, it's an appropriate time in our lecture series as we end section 26, we'll lecture 26, is to review how do you take the derivative of various exponential expressions? That is, you have a base, you have an exponent, what do you do when you wanna take the derivative? Well, if you're gonna take the derivative with respect to x in all of this situation, we'll just do it with respect to the variable x. What if you take the derivative of the term A to the B, where A and the B are both constants? They're just numbers. So you're taking the derivative, for example, of something like e to the seventh. Well, it's very tempting for people to be like, e to the seventh, well, by the power rule, that's equal to seven e to the sixth. Well, that's not right. Some people are like, oh, it's just e to the seventh, the derivative of e is just itself. Well, that's not true either. The issue here, and the thing that we get confused with is derivative of e to the x is equal to e to the x, that's very true. So the exponential function is equal to its own derivative, but we don't have that here. This is a constant. This is just a number. e to the seventh is a number. It's derivative of zero, much in the same way that the derivative of five is equal to zero. So if your exponential expression, A to B is just constants, the derivative of zero, it doesn't matter if it's an exponential expression. If it's a constant, it'll be zero there. Well, okay, what if the base is a function? There's some variability to it, but then the exponent is still a constant. Well, in that situation, then the power rule comes into play. The power rule is what you would use, and depending on how complicated f of x is, you might have to use the chain rule as well. So this would be something of the form. Let's say we have like an x squared plus one, and we raise this to the three fourth power. If we want to calculate the derivative of such a thing, by the chain rule, we'll get three fourths times x squared plus one. We lower the power by one to the negative fourth power, and then we multiply by the inner derivative, which is a two x, and you can simplify there from there. But when your base is a variable, it's a function, but the exponent is a constant. In that case, you just use the power rule, no big deal. Well, what if we switch the roles? What if the base this time is constant and the exponent is where the variable is? That's a function. So this would be something like take the derivative, let's say of E to the x squared plus one power. What if you had something like that? Well, in that situation, the chain rule comes into play, but you're also gonna use the fact that we can take the derivative of x exponentials. The derivative of the natural exponential is itself. So you just get E to the x squared plus one, and then you times that by the inner derivative, which is a two x right there. And if your base was some other base, like base two or base three, it would be essentially the same calculation, but then you have the times by the natural log of two and the natural log of three, whatever that base, whatever the base is, you have to pay the tariff there. So just treat it like an exponential function using the chain rule. The last case which we actually haven't explored yet in this lecture series is what if the exponent and the base have variables? What if they're both functions that vary in their own right? You can't use the power rule because the power rule requires a constant exponent. You also can't treat it like an exponential function because that requires a constant base. But it turns out that all four of these creatures, all four of these expressions can be, the derivative can be computed using logarithmic differentiation. In fact, we prove the power rule using logarithmic differentiation. The power rule is just a special case of logarithmic differentiation. So in this fourth case where the base and the exponent are variables, they're functions, you pretty much have to use logarithmic differentiation. There is a general formula you can use here, but it's not one I would recommend memorizing. I think it's just too easy to make a mistake. Consider the function right here, y equals x to the sign of x. Notice that the base is a variable x as x changes, of course. But the sign of x also will change as x changes. So what do we do? Well, what we're gonna do is we're gonna use logarithmic differentiation. Take the natural log of both sides. So we get the natural log of the absolute value of y equals the natural log of the absolute value of x to the sign of x right here. Now the advantage of taking natural logs is you can take the exponent out of this thing. So we end up with sign of x times the natural log of the absolute value of x. Now take the derivative of both sides. So take the derivative with respect to x on both sides. Well, by the chain rule, when you take the derivative of the natural log of y, you're gonna end up with a y prime over y. On the right-hand side, we're gonna have a product of two things. So use the product rule. We're gonna take the derivative of sign times the natural log of the absolute value of x, and then you're going to add to that sign of x times the derivative of the natural log of the absolute value of x right there. Take its derivative. Derivative of sign is a cosine of x times the natural log of the absolute value of x. And then the derivative of the natural log of the absolute value of x, that's gonna be a one over x, and this is all equal to y prime over y. So the next thing you wanna do is multiply both sides of the equation by y, y, of course, being x to the sign of x power. So this gives us that the derivative of y prime is going to equal cosine of x. Well, excuse me, I forgot the y there. We have to multiply both sides by y. y here, so the cancel, multiplied by the y here and distributed through. Well, actually we'll just leave it factored. y, of course, is x times sign of x right here. And so then you're gonna get cosine of x times the absolute, or times the natural log. Well, we put the absolute value of x here, but we actually can drop the absolute value of x here because the original function, x to the sign of x power, the issue here is with these exponential expressions, if we allow our base to be negative, then you potentially could get some imaginary numbers. So when you see something like y equals x to the sign of x power, it goes without saying, but I probably should say right here anyways, that the domain of this thing is gonna be greater than or equal to zero. Negatives aren't allowed because that could give us some imaginary numbers. So because x is necessarily greater than or equal to zero, if we didn't put the absolute value in there, that's okay, because the absolute value of x will just be x, because it necessarily is non-negative here for this function. So we get cosine of x times the natural log of x plus sign of x all over x, which then gives us the derivative of this function for which we had to find it using logarithmic differentiation. There's no other technique that we've used in our lecture series that would prove to be more effective than just logarithmic differentiation right here.