 OK. Welcome to a screencast about the derivatives of constant multiples. So the constant multiple rule is used whenever you have a constant that's being multiplied by a function. So this says for any real number k, if f of x is a differentiable function, so that means you can find the derivative of it. And it has a derivative that's defined to be f prime of x. Then if we want to do the derivative with respect to x of k times f of x, that's simply k times the derivative f prime of x. OK. So find the derivative of each of the following functions using proper notation. So we're given m of x is equal to 4 times x squared. So the 4 obviously is a constant that's being multiplied. So in this case, that's our k. So now we've got to remember back, how do we do the derivative of x squared? Well, x squared is a power function. So we're going to want to follow the power rule. So m prime of x is going to equal 4 times the derivative of x squared is 2x. And that's also better known as 8x. So kind of once you get the hang of these, a little bit of a shortcut when you have a power anyway, is you can just multiply the power by that constant out front and then reduce your power of your exponent by 1. OK, h of a equals 10 times the square root of a. Now the square root, well, that's not really anything that we know right off the bat for any kind of rules. But I can rewrite it as a power and then use the power rule. So that's going to be 10 times a to the 1 half power. So this is very similar to the one we just did up above. So h prime of a is going to equal, bring down that coefficient 10 times. Now we need to do the derivative of a to the 1 half. Well, again, that's like a power x to the n. So we can follow our power rule. So that's going to be 1 half a to the negative 1 half. And again, the more you practice these rules, hopefully they'll just become second nature. And you can just look at these and do them. But you have to practice to get to that point. So if you do want to simplify this a little bit, we can go ahead and multiply that 10 by the half if you wanted. And that would give us 5a to the negative 1 half power. And again, you can do more with this exponent being negative in a fraction if you want to. But that's certainly up to your instructor. OK, next example. I want to determine d over dt. So that means the derivative with respect to t of negative 2 times 4 to the t power. So my negative 2 is my coefficient. So that's going to be like my k. But this one is 4 to the t. So is this a power function or is this an exponential function? Those are the two we've looked at so far. And since the variable's in the exponent, this is definitely an exponential. So to do the derivative of this, we're going to have to follow our exponential rule. OK, so my negative 2 just comes along for the ride. And remember, 4 to the t, again, comes back at it. So 4 to the t. But then we have to multiply by the natural log of the base, which is 4. Now the ln of 4 is a number. Negative 2 is obviously a number. Those are all just your coefficients. So again, if you want to simplify this a little bit further, you can. But I certainly think this answer is perfectly good enough. Thank you for watching.