 Hello and welcome to a screencast about the derivatives of exponential functions. So an exponential function, if you recall, is of the form a to the x. So a is a positive real number, and that's called your base. And x, then, is your variable, which is in the exponent. If you have a function of this form, then the derivative looks like a to the x. So that comes right back at you. But then you're going to have to multiply it by the natural log of your base, which is a. Okay, so example one, we want to look at a few derivatives here. And we want to find the derivative of each of the following functions, again using proper notation. So g of x equals 2 to the x power. That definitely has this form because the x is in the exponent and 2 is a positive real number. So in this case, our derivative is going to look like the same function coming back, 2 to the x, and then times the natural log of 2. Right, next one we have p of t equals 1 fifth to the t power, okay? So again, this has the same kind of a form to it. It's in the form a to the x, or in this case, a to the t. Where a is 1 fifth, 1 fifth is definitely a positive real number. So we can follow the same formula. So that's going to be 1 fifth to the t. Again, that comes back atcha. And then times the natural log of 1 fifth. Right, next example, I have this different notation again. So this is d over dz of e to the z. So remember that d over dz just means take the derivative with respect to z. So in this case, z is our variable. Now, you may also be looking at this one and see an e. But remember, e is actually a number. It's like 2.7, whatever. It's definitely a positive real number. So we can, again, just follow the same rules as above. So that's going to be e to the z. That doesn't change. And then that's going to be times the natural log of e. Now, this one's a little bit special because remember, the natural log and the e function, they're inverses. So they're going to cancel. So in this case, the derivative of e to the z is just e to the z. Thank you for watching.