 Hello, and today we are going to be talking about doing derivatives of sums. So the sum rule as I defined here says that if f of x and g of x are differentiable functions and their derivatives are f prime and g prime respectively, then the derivative with respect to x of f of x plus g of x is simply as kind of your intuition would follow, it's f prime of x plus g prime of x. So again, remember this d dx notation just means derivative. Okay, example one says find the derivative of each of the following functions using proper notation. So the first function here we have is a polynomial and that polynomial is p of x equals 3x to the 10th minus 4x to the 6 plus 15x minus e squared. Okay, so it would be possible to do this the long way that we used to do it using the limit definition, but now that we are using all of our different rules, we can go ahead and apply those. So p prime of x equals 3x to the 10th. So that's one of those constant multiple rules, and then in here I see a power rule. So we can go ahead and multiply the 3 by the 10 and get 30x to the, reduce the power by when you get 9, minus. Now I know this is just called the sum rule, but remember minus is very, or subtraction works very similar to addition. So this would be minus, then we multiply the 6 by the 4, and we get 24x to the 5th plus the derivative of 15x, which is just 15. An easy way to remember that derivative is 15x, if you were to graph it, it's just a line that has a constant slope of 15. You could also remember this is x to the first power. So then when you multiply the 1 by the 15, that will give you a 15, and then reduce that power by 1, that gives you 0. But anything to the 0 power is just 1, so your answer there is just 15. And then now the last piece here can be a little bit tricky. So e squared, you could just type that number on your calculator or in any other device, and you're gonna get a number out of that. So in this case, that derivative is gonna be 0, because e squared is just a constant value. So that's your derivative, and then you're done. Okay, the next one I wanna look at is g of t equals 2 minus 6 times the cube root of t plus 5 to the t minus 17 over t to the 8th. Okay, so this one I'm starting to mix up a few different rules that we have. So I see a root in here. So I see the cube root of t, we can probably do something with that. I see, first of all, I see a constant 2 that we know what happens with that. 5 to the t, that's an exponential, okay? So we're gonna have to apply a different rule to that one. And then 17 over t to the 8th here, again, we can do some mess here. We can do some algebra to simplify that messiness. Okay, so first I'm gonna do some rewriting. So 2, we're just gonna rewrite minus 6. Now thinking back to your algebra days, the cube root of t is the same thing as t to the 1 third power plus 5 to the t. Cuz that one really can't be rewritten anywhere. And then again, thinking back to your algebra days, 17 over t to the 8th is the same thing as 17 t to the negative 8th power. Okay, so now we can do just like we did up above, go through and take the derivative of each piece. Again, you've gotta remember though, are we doing a power? So we can do the power rule, or are we doing an exponential? So we can do the exponential rule. And then just Morello's coefficient just come along for the ride. So the derivative of 2, that piece goes away, okay? Because 2 is a constant, so the derivative of that's gonna be 0. So we have negative 6 t to the 1 third, okay? Well, this is just a power, so we can drop that power down. So negative 6 times 1 third is negative 2. t to the 1 third minus 1, so that's 1 third minus 3 thirds. So that's gonna give us a negative 2 thirds plus, okay? So that's our power rule. 5 to the t though is an exponential. So we're gonna have to use the exponential rule. That means the 5 to the t comes back, and then we have to multiply it by the natural log of our base, which is 5. And then let's see, we've got negative 17 times negative 8. I guess I should have grabbed my calculator for this one. So the arithmetic for that is gonna be a positive. So I can do this one in my head, 136. I better double check the math there though. t to the, so let's see, negative 8 minus 1 gives us a negative 9, okay? Fantastic, okay, then the last one here, and again, I just wanna continue to use this notation so you guys get familiar with it. d over dx, so that just means the derivative with respect to x of our function, negative 3 e to the x plus 2 over the fourth root of x, okay? So e to the x, that's an exponential, cuz I see a base in there and that's actually a number, even though e looks like a letter. And then again, we're gonna wanna do some rewriting with the second piece. So let me do some rewriting first. This is where it's always smart to take a step to rewrite it and then take a different step to do the derivative. You don't wanna start jumping into a derivative of one piece and then rewrite the second piece cuz then you're just gonna confuse yourself as to what's going on. So this is negative 3 e to the x plus, let's see, this will be two times. So the fourth root of x is the same thing as x to the 1 fourth power. But the fact that it's in the denominator means you can bring it up and it becomes negative, okay? So there's some algebra for you. Now we can go ahead and apply our rules that we've been learning with calculus. So negative, the derivative of negative 3 e to the x, the negative 3 comes along. The e to the x comes along. And then technically, we'd be multiplying by the natural log of e. But remember that's just one. So the derivative of negative 3 e to the x is just negative 3 e to the x. Don't make that one any harder than it really is. Okay, now we have to multiply 2 by negative 1 fourth cuz this is a power. So this will be negative 1 half x to the negative 1 fourth minus 1. So negative 1 fourth minus 4 fourths gives us negative 5 fourths. Okay, thank you for watching. Oops, that's a four. Thank you for watching.