Question

The oracle function f(x,y) is presented below. For each point (x,y) you enter the oracle will tell you the value f(x,y). Estimate the partial derivative of the function at (1.6,−0.0999999999999999) using the Newton quotient definition.

Answer 1

you need a difference. if you are looking at the given point, try these points into the oracle x=1.59, and x=1.61 (h=.02), and you now have f(x+h), and f(x)
then it will give you f(x) at those points, then do this https://www.youtube.com/watch?v=1O5NEI8UuHM

Answer 2

Suppose you want ∂f/∂y
We can approximate that by
(f(x,y+h)-f(x,y)) / h
So just plug and chug. For ease of typing, I'll use (1.6,-0.099)
You can add as many zeroes as you want.
(f(1.6,-.1+0.001)-f(1.6,-0.1))/0.001
You have not said what f(1.6,-0.1) is, so that's all I can say at the moment.

Answer 3

I just did that exercise on my webwork assignment, do the following:

1) Calculate in the oracle f in the point 1 they gave you (1.6,−0.0999999999999999). Let's call this result f(x1,y1).

2) Calculate in the oracle f in a point slightly superior in the x axis. For example (1.601,−0.0999999999999999). Let's call this result f(x2,y1).

3) Use the Newton's quotient definition:
Fx (x1,y1,) = ( f(x2,y1) - f(x1,y1) ) / (x2-x1)

Where x2 is 1.601 and x1 = 1.6.


Be happy :)