As I promised, two more examples of partial derivatives. In this case, these two examples should do a lot to solidify some, especially foundational concepts that you should have brought to this level of the course. If not, it's a good time to re-evaluate a few things. One is, first of all, this function h is in terms of s and t. If that confuses you, if you're only comfortable dealing with functions that have x, y, or z in them, then you need to start thinking that a function is just an input creates an output. The variable names themselves don't matter. So h of s and t is just a function of s and t. You could rewrite this f of x, y if you really wanted to, but being comfortable with functions of different variables at this point in the course should be something that you're getting accustomed to hopefully. We also see natural log in here, we also see a root that is not a square root. Maybe you haven't seen that before. Maybe you have. We see another trig function, cosine, and we see e to a power, and we're going to figure out how to take the derivative of that. If all the things I already said, you really know these concepts, that's great. If you don't, it's just time to, one, learn them right now, and two, go back maybe and practice a bit more if you feel you need it. Partial derivative of h in terms of s is, well, if I'm taking the partial derivative in terms of s, then t is a constant, so t^7. The derivative of the natural log of s squared is 1 over s squared times the derivative of the inside, which is 2s. Then 9 divided by t^3 is just a constant, so that goes away, 0. Then here we are. Let's try to do this. S^4 and then the seventh root of that is equivalent to saying s^4/7. These are just fractions and exponents. Again, something you should feel confident about going from here to here or here to here. So minus 4/7s to the 4/7 minus 1, so negative 3/7. That is t^7 times 2s over s squared minus 4/7s to the negative means 1 over the seventh root of s^3. S^3, seventh root, the negative implies that it should be on the denominator unless it was already on the denominator, which goes to the numerator. This is just algebra of a negative and an exponent is again a concept that takes a little bit of practice. Partial derivative of h in terms of t will give me, so natural log of s squared is now considered a constant or treated as one at least, so we just take the derivative of t^7 which is 7t^6 times ln of s squared plus, now nine t^1/3 can be thought of as a 9t to the negative 3. We just bring over the negative 3. Negative 3 times 9, 27t to the negative 4. Then this one goes to zero. It would be 7t is 6 natural log of x squared minus 27 divided by t^4. Blue for this one, let's do this one. Again, this one becomes a little bit more complicated. Let's do f of y first because that one is definitely more straightforward. The partial derivative of f terms of y is, cosine of 4 over x is a constant, so that stays. A derivative of e to a function will be e to said a function times the derivative of that function. Still assuming that y is the one that you're taking the derivative in terms of x as a constant. In this case, x squared y in terms of y will give me x squared, because that's treated as a constant, minus 3 times 5 is 15y squared. Now let's do partial derivative of f in terms of x. Now, here you need to realize that this is two functions of the variable that I'm taking the derivative in terms of. Here it's x. Function of x times a function of x, we will need to use the chain rule here. We'll take the derivative of the first times the second plus the first times the derivative of the second. The derivative of cosine of 4 over x is negative sine of 4 over x times the derivative of the inside. You can think of 4 over x, that just equals 4x to the negative 1. If we take the derivative of that, it's negative 4 over x to the negative 2. That's the derivative of the cosine part alone. The derivative of the first times the plain old second, which is x squared y minus 5y^3. Again, chain rule comes up. If you're not confident with the chain rule, you're definitely not going to be confident with the chain rule inside of partial derivatives because partial derivatives can get even more complicated than derivatives that need the chain rule with just one variable. Now, plus the original times the derivative of the second, which is e to the function times, remember the derivative of this is just this times the derivative of the top in terms of x. In terms of x will give me 2xy. Hopefully this solidifies some of the ideas of partial derivatives. Of course, that's our goal here. But just like a lot of things in math and in data science, if you don't understand the big foundational concepts, then moving forward is often difficult. If you don't understand exponent's or negatives of an exponent or fractions in an exponent or natural logs or trig, then sometimes it can catch up to you. These four examples we've done, reflect on them a little bit, see how you feel about them, maybe try some other examples, and then if needed, you can always stop, go back and do a little bit of extra practice.