Mother in this problem were asked to find all these second order partial derivatives of this functions. Let's begin with the first order partial derivatives W sub X. So treating excellent the variable here we can tell we're going to need the product rule since we have one function of X times, another function of X. So buy the product will we're going to need the first function X times the derivative of the second function. By derivative, we mean the partial derivative with respect to X of this second function.
Okay, plus the derivative of the first function times the second function. That's just the product rule that we know and love. Let's go ahead and compute this. What's the derivative of it? Sign of X squared y. Well, we'll start with the chain rule here, since we have a sign of some function.
So starting from the outside and the derivative of Sinus co sign, we keep the inner function here and we have to multiply by the derivative of this inter function again. When we say derivative, we mean partial derivative with respect to X. So since why is a constant the derivative of this inter function here is to x y that the wife stays there and derivative exporters to X to make sure you're convinced. Event moving on the partial derivative of X with respect to X is just one, and the rest can stay there. So cleaning up her answer just a little bit the it to X squared.
Why co sign next squared y plus sign X squared Y all right w sub y. It will be a little easier because this time X is treated as a constant, so we don't need the product rule. That X is just a constant multiple. So that stays there, the derivative of Sign, He said his co sign, the Inter Function stays. Now we multiplied by the partial derivative of X squared y with respect to why which is just X squared.
So you end up with X Cubed co sign X squared y all right, so we'll circle those just so we know where to look these air first, our first order partial derivatives. Now time for the second. Let's start with W sub x x. So we have another product rule to do so. Make sure you can follow along each step, feel fear to pause the video if you need to, at any point product tool for this 1st 1 just so we know where we are.
We're just taking care of this first term right now. So notice it's foots into two parts. The two X squared y and the co sign of X Squared y. So how if we go, product rule tells us we need the first function to explain why times the derivative of the second with respect acts the derivative of coastline is negative. Sign of X squared y and the derivative of X squared y with respect to X is we said to X y all right, plus the derivative of the first function with respect to X.
So the derivative of two X squared y is four x y Make sure you're convinced of that times the second function co sign of export Why, All right, so we're not done. It's just that was the product rule of this first blue parte eso underlined this in blue. So we're halfway there. We still have to add I'll make it red, the derivative of sign of X squared y. This will be a little faster so the derivative of Sinus co sign.
We keep the inter function now we multiplied by the derivative of X squared y with respect to X, which is to x y. Okay, so let's clean this up. Um, so what do we have? Let's see, this whole first term two times two is four. We do have a negative sign in there, so we should be getting negative for extra third y squared sign of X squared Wise all comes from the very first term, and now we have notice we have four x y coastline X squared y plus two x y Costa Next chord. Why those air actually like terms So the four on the two can add.
We end up with six x y co sign of X squared y. That's about as far as we can go. I think so. That is tthe e answer for wciv xx w. Some while the little quicker because we don't need a product rule in that X to the Third Street is a constant again Derivative of coastline is negative.
Sign. Let's put the negative out front. It's sign that inter function X squared y now times the derivative of X squared y with respect to why which is X squared. So we end up with minus X to the fifth sign of X squared. Y ok, getting there now.
W sub X y this would really be a pain again, huh? To go Thio Tate to take w x and take the partial derivative with respect a wife, we would again need a product rule here. Um, so it's clever to say, Wait a second. Look at all of our functions that we've found so far the original and all of our derivatives. They're all continuous, all they ever polynomial ls and signs and co signs and all those air continuous And so we can invoke Claros. The're, um there.
I'm too from this section and assert that w x y actually equals W Y X. And in this case, it's smarter to do it this way because, um, this should save us a little time, right? You will still need a product rule if we're looking at, uh, taking the derivative of this down here with respect to X. But at least there's only one term and not too. So it'll save us a little bit of time So again we have a product to A We have a first function in a second function. We're taking the derivative with respect to X.
So the first function next to the third. Now we want times the derivative of the second part, the derivative of co sign is negative Sign uh, X squared y times the derivative of X squared y with respect to X, which is to X y Kim and the second half of the product rule Now times the derivative of the first part. That's three x squared times a second function. Okay, and we can rewrite that down here. So let's see what we're going to have.
The first term will be negative, and we'll have minus two X to the fourth. It looks like exit the third times X. Yep, we do have a Y in there, and it's a sign of X squared y, and in the second term is a simplified as it can get those three x squared co sign X squared. Why, all right, we'll put a little boxer on that, and we're done because that was both of our mixed partial derivatives equals this. So we have found all four.
Hopefully, that helped.