5

FindI6xy) Ly(xy} Ix6xy} and {y(ky) for the following fncton Sx 10y fxy) = Ty 3x{xky)-[L...

Question

FindI6xy) Ly(xy} Ix6xy} and {y(ky) for the following fncton Sx 10y fxy) = Ty 3x{xky)-[L

FindI6xy) Ly(xy} Ix6xy} and {y(ky) for the following fncton Sx 10y fxy) = Ty 3x {xky)-[L



Answers

For the following exerises, find $\frac{d y}{d x}$ using parial derivatives.
$$
x^{3}+y^{2} x-3=0
$$

Okay, So for this problem were given that X to the two thirds plus, why did the two thirds equals a to the two thirds? But I want to rewrite. This is being X to the two thirds plus, why do the two thirds minus eight of the two thirds equals zero? So now what we want to do is for our formula. We're going to be finding d y over DX. So what I want to do is create a negative fraction of D F, which is gonna be my rich, my equation over de X and then d f over d y. Okay, so negative fraction. So I'm going to take the derivative of this in terms of X. So this is going to end up being two thirds X to the negative one third over, um, And so then plus, why do two thirds is a constant and minus eight to the two thirds is a constant. So then I'm gonna look at the drive of in terms of why so now X to the two thirds is a constant. I'll be left with two thirds. Why? To the negative one third. And then, of course, a to the two thirds is a constant. So if I notice the two thirds have something comments, Aiken, get rid of these, and then I'm gonna have minus since both of these exponents or negative, I'm going to flip them, and this would be my answer.

This problem. Once you find dy Of Y is equal to three X -10. Five. Before we can find Dy we need to find the derivative Dy over dx. So the derivative would be the extended product rule, meaning we would have five times three X minus 10 to about four times The derivative of this group in here, which would be three. So there's that derivative. That's the Dy over dx. That's f prime of X. So dy would then equal five times three X minus 10 to apart four times three Times DX. Which is given the question at 0.03. Now, it also says that X is equal to four. So I'm going to race the X and plug in a four and then let's calculate that out. So do I would be equal to five times 12 minus 10 is two to the power four times three times zero point 03 And we get two times two is four times two is 8, 10 to 16. 5 times 16 times three times 0.3 And you get an answer when you multiply that out of 7.2

Execute plus two X and dX DT is five. Find DY DT when excess too. All right. So, first, we'll just take the derivative of this equation with respect to T. So the derivative of Y is DY DT. That's okay. And then the derivative of X cubed three X squared times the derivative of X, which is dx DT Plus two times the derivative of XDX DT. So everywhere is the X. Put it to everywhere. See dX DT five. So three times two squared times five Plus two times 5. So that is uh, five 1020s, 2060 plus 10, Which is what she calls 70.

So we need to find the derivative of the function. Why is equal to X plus three times X minus two. So let's foil this and we get X squared minus two X plus three x minus six. So this gives why is equal to X squared, plus X minus six. So now the derivative of why, with respect to X is equal to you two. X plus one.


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