Partial Differentiation - The Chain Rule

[krackedude]

Homework Statement

Calculate ∂f/∂s + ∂f/∂t at s = 2, t = -1.

Given:
f = f(x,y)
x = s - t
y = s² + t²
∂f/∂x (3,5) = 0.06170
∂f/∂y (3,5) = 0.06170

Homework Equations

∂f/∂s = ∂f/∂x * ∂x/∂s + ∂f/∂y * ∂y/∂s

∂f/∂t = ∂f/∂x * ∂x/∂t + ∂f/∂y * ∂y/∂t

The Attempt at a Solution

I calculated:

∂x/∂s = s = 2
∂x/∂t = t = -1
∂y/∂s = 2s = 4
∂y/∂t = 2t = -2

I know I haven't gotten far, but I don't know how to get ∂f/∂x or ∂f/∂y from the given equations.

 

[SammyS]

What is are the values of x & y when s = 2 and t = -1 ?

 

[Dick]

They gave you ∂f/∂x and ∂f/∂y at the point (x,y)=(3,5). What might the point (x,y)=(3,5) have to do with the point (s,t)=(2,-1)? And I don't think ∂x/∂s = s = 2 is right either, recheck that and ∂x/∂t.

 

[Mark44]

Homework Statement

Calculate ∂f/∂s + ∂f/∂t at s = 2, t = -1.

Given:
f = f(x,y)
x = s - t
y = s² + t²
∂f/∂x (3,5) = 0.06170
∂f/∂y (3,5) = 0.06170

Homework Equations

∂f/∂s = ∂f/∂x * ∂x/∂s + ∂f/∂y * ∂y/∂s

∂f/∂t = ∂f/∂x * ∂x/∂t + ∂f/∂y * ∂y/∂t

The Attempt at a Solution

I calculated:

∂x/∂s = s = 2
∂x/∂t = t = -1
∂y/∂s = 2s = 4
∂y/∂t = 2t = -2

No, this isn't right.
x = s - t,
so ∂x/∂s = 1 and ∂x/∂t = -1

y = s² + t²,
so ∂y/∂s = 2s and ∂y/∂t = 2t

Now, when s = 2 and t = -1, what are the values of the four partials?

You are confusing a function (such as ∂x/∂s = s) with its value at a particular number in its domain (such as ∂x/∂s(2) = 2).

I know I haven't gotten far, but I don't know how to get ∂f/∂x or ∂f/∂y from the given equations.

You don't need the formulas for ∂f/∂x and ∂f/∂y. You are given the values of these functions at the point (3, 5) in their domains.

 

[krackedude]

Wow...I can't believe I missed that one. x = 3 and y = 5, so the given values can be used to calculate it. Thanks a lot!

Edit: yeah, i definitely screwed up ∂x/∂s, and ∂x/∂t...Thanks!