Calculating Change and Derivatives for a 2 Variable Function

[hoeranski]

Homework Statement

f(x,y) = (x-y) / (x+y)

Calculate Δf and df for the change of point (3,2) to point (3.2, 2.1)

Homework Equations

The Attempt at a Solution

I guess that I have to use limits, but don't know how to begin with.

 

[HallsofIvy]

Are you saying you do not know what \Delta f and df mean or are you simply unable to differentiate?

\Delta f= f(3.2, 2.1)- f(3, 2)
df= \frac{\partial f}{\partial x}dx+ \frac{\partial f}{\partial y}dy

 

[hoeranski]

Yes, I know what they mean, but I didn't know where to start.

So Δf should be:
Δf = (3.2 - 2.1)/(3.2 + 2.1) - (3 - 2)/(3 + 2)
Δf = 1.1/5.3 - 1/5
Δf = 0.007547...

And df should be:
df = ((2y)/(x+y)^2)*dx + ((2x)/(x+y)^2)*dy

Are my attempts correct ?

 

[epenguin]

Yes, I know what they mean, but I didn't know where to start.

So Δf should be:
Δf = (3.2 - 2.1)/(3.2 + 2.1) - (3 - 2)/(3 + 2)
Δf = 1.1/5.3 - 1/5
Δf = 0.007547...

And df should be:
df = ((2y)/(x+y)^2)*dx + ((2x)/(x+y)^2)*dy

Are my attempts correct ?

I think the first part is right. With hindsight you shouldn't see it anyhow difficult - there is the function, and the change in the function is the change in the function - the difference between its values for that x,y and this x,y.

The second part is not part of the question but good to do. I think your second term has a mistake of sign. To see this don't just do the differentiation again, but notice the symmetry, that f(x, y) = - f(y, x) . Such symmetries are often useful checks for errors and shortening calculations.

Then if you put in place of dx, dy Δx, Δy you ought to see whether you get a fair approximation to the previous result, though it will not be exact (except by accident).