# 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 ??

## Answers and Replies

BvU
Science Advisor
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
Science Advisor
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.