# Total differential Problem

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

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
Homework Helper
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.

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

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

Oh I see, thank you very much
You're welcome.

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.