How Does Total Differential Determine Accuracy in Parameter Settings?

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The discussion revolves around using the total differential to assess parameter accuracy in the function z(x, y) = 3sin(x² + y)y + x³. Participants clarify the calculation of the partial derivative ∂z/∂y, specifically addressing the inclusion of zero and one in the expression. The zero arises from the derivative of x³ with respect to y, while the one is a result of applying the chain rule. There is also a reminder about forum etiquette, emphasizing the importance of allowing learners to engage with the material rather than providing direct answers. Overall, the conversation highlights the nuances of calculus and the importance of fostering a learning environment.
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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 ??
 
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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.
 
Megaquark said:
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.
 
BvU said:
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
 
Megaquark said:
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
 
Nanu Nana said:
Oh I see, thank you very much
You're welcome.
 
BvU said:
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.