Let's start off with two examples. One is slightly easier and then one where we'll need a little bit of trig knowledge and get into a few more higher up concepts on this one. But let's start off with this one. Here we have x and y are in the function, so we know f of x, y equals z. We're just calling it z for convenience sometimes, and we want to take the partial derivatives of this. Let's do partial derivative of f in terms of x. Now, we talked about last time that if you want to take a partial derivative in terms of a certain variable, all the other variables need to stay constant in concept. They're going to stay as y, they're not going to turn into a constant or anything like that, but we're just treating them like we would a constant. When you take the derivative of a constant alone or being multiplied by another constant or something like that, you'll get zero. Keep that in mind here when we go forward. Our first term is x^4, so we want to take the derivative of that in terms of x, we'll get 4x^3. Now, we get to 6 times the square root of y, so keep in mind that 6 times the square root of a constant is just a different constant. When we take the derivative, we'll get 0 and 10, it's just a constant. When I take the derivative, we'll get 0. You just get 4x^3 here. Let's do partial derivative of x terms of y. In this case, we're taking the partial derivative of f in terms of y, so x will be a constant in concept, a constant to the fourth. You just separate this a bit, we know what we started with. The derivative of a constant to the fourth is just 0. If you had 2^4 there, if you take the derivative, you would get 0. Now, keep in mind, if the square root get to you and they're tough and concept for you, you can think of this as 6y^1/2. If you take the derivative of that, you'd get 1/2 times 6 times y^ negative 1/2. We bring the 1/2 down times 6 times y, subtract 1 from that 1/2, so 1/2 minus 1 will give me negative 1/2. At this point, this term itself, when you take the derivative in terms of y, that's just a normal derivative. If these little derivatives are confusing you, then you should go back and review general concepts of derivatives. Then minus derivative of 10 in terms of y would be 0 as well. We'll get 3, 1/2 times 6 will give me 3, and then y^ negative 1/2. Again, concepts of exponents is are really important, then negative will bring me down to the denominator, the 1/2 will give me square root. We'll have 3 divided by the square root of y for that. This is again df/dy. We did our first partial derivative, great. Let's go on to something a little bit more complicated, not too crazy. It will use trig if trig was not really your thing and you didn't put too much time into it, it's okay. You'll learn one of the derivatives of the 10 function today. Again, if it confuses you or you have no idea where it came from and you haven't put time into true, that's completely fine. It's up to you whether you want to go back and learn the basic trig functions and their derivatives, your call. Here we want derivative of w, which again is a function of x, y, and z. Derivative of w in terms of x. Now, we're taking x squared times a constant. The constant will stay there, so while the stay, derivative of x squared is 2x and y it's just a constant, so it'll stay. There's no x in here, so 10 times a constant squared times a constant to the third, it's just a different constant. Derivative of that is 0 plus 43x, derivative of that in terms of x is just 43. Then this is only in terms of y, so that's a 0. We get 2xy plus 43. Making some good progress here. Partial derivative of w in terms of y. In this case, y is our variable, x squared is our constant. You can think about x squared y. What would you do if you took the derivative in terms of w of 3y? The y would just go away, you'd end up with 3. Here x squared y, x squared is just a constant, the y would go away, you end up with x squared. Here we have y squared, z^3 is a constant, 10 is a constant, so that just stays. Derivative of y squared is 2y. So 2 times 10 is 20yz^3, plus 0 because 43x is just a constant minus 7. Now y is in here, so the derivative of 10 of a constant times y will give me 7. The original 7 is there, derivative of 10 is secant squared. Then you have to remember to multiply by the derivative of the inside, which is simply 4. We will get x squared minus 20yz^3, the 0 will go away, minus 4 times 7 will give me 28 secant squared for y. In terms of z, we can do that. Hopefully we have room to do that down here. I'm not sure if we do. It's really straightforward. The only term with the z is this one. Z^3, derivative will be 3z squared, so z squared here. The 3 comes down times 10 is 30, and the y squared is just treated as a constant. It's 30y squared z squared. Partial derivatives are new concept. Maybe your derivatives weren't too strong, maybe your trig isn't too strong, maybe your exponent algebra isn't too strong. Hopefully this is a good start. Next time, we're going to do two more examples of partial derivatives. If you feel you need it, feel free to tune in. If you feel you don't need it, feel free to not tune it. But we're going to do that next time, and then after that, we'll go on to a new concept.