Total differential Problem

[Nanu Nana]

Homework Statement

You have two parameters x = 12 and y = 3 set on a machine. The machine generates a function: z (x, y) = 3sin (x ^ 2 + y) y + x ^ 3
Use the total differential of this function in the set point to determine which of the parameters to be set to the most accurate.

Homework Equations

dz = (∂z/∂x) dx + (∂z/∂y) dy
3.Solution
z = 3y sin(x²+y) + x^3
[/B]
dz = (3y cos(x²+y) * 2x + 3x²) dx + (3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy
dz = (6xy cos(x²+y) + 3x²) dx + (3 sin(x²+y) + 3y cos(x²+y)) dy

I don't understand why (∂z/∂y) = (3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy Where did that zero came from ? and 1 ??

 

[BvU]

So, NN, where is your attempt at solution ?
What would be your $\partial z\over \partial y$ ?

 

[Megaquark]

Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x³ with respect to y. The 1 comes from the y...chain rule.

 

[BvU]

Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x³ with respect to y. The 1 comes from the y...chain rule.

Did you notice the posting in 'homework' ? It is not good for the poster and it is against PF rules to give such a direct answer: it robs the poster from an opportunity to learn from insight.

 

[Nanu Nana]

So, NN, where is your attempt at solution ?
What would be your $\partial z\over \partial y$ ?

(3 sin(x²+y) + 3y cos(x²+y) * 1 + 0) dy

 

[Nanu Nana]

Writing he zero is kind of unnecessary, but it comes from taking the partial derivative of x³ with respect to y. The 1 comes from the y...chain rule.

Oh I see, thank you very much

 

[Megaquark]

Oh I see, thank you very much

You're welcome.

 

[Megaquark]

Did you notice the posting in 'homework' ? It is not good for the poster and it is against PF rules to give such a direct answer: it robs the poster from an opportunity to learn from insight.

Nope, I didn't notice. I'll likely avoid answering posts in this section from now on.