[Note that, in thisexample, the mesh sizes in x and y are identical (h); strictlyspeaking, this need not be true. In some applications, we may needmore resolution along the x- or y-axis; we could then use separatemesh sizes h_x and h_y.]

By definition, thepartial derivative of a function f ( x , y ) with respect to xis

∂ f ∂ x = L i m h ⟶ 0f ( x i + h , y j ) − f ( x i , y j ) h

and the partialderivative with respect to y is

∂ f /∂ y = L i m h ⟶ 0f ( x i , y j + h ) − f ( x i , y j )/ h

If we applied theseformulas to our grid values, we would get the finitedifference expressions

∂ f /d x ( x i , y j )≅ f ( x i + 1 , y j ) − f ( x i , y j )/ h

Note:To avoid round-off error, retain at least six decimalplaces in all of your calculations.

• Assume the function f is defined as f(x, y) = 3 tan x cosy
• Use differentiation rules to find the exact partial derivatives∂ f /∂ x and ∂ f /∂ y , and evaluate those exact partialderivatives at (1.56, -2.1).  
• Use the finite difference formulas to estimate ∂ f /∂ x and ∂f/ ∂ y at (1.56, -2.1) for three different values of the mesh size
  • h = 0.01
  • h = 0.001
  • h = 0.0001

• Use your calculated values to fill in this table:

Estimated partialderivatives using finite difference formulas:
h	finitedifference approx. to ∂ f/ ∂x	exact ∂f/ ∂ x	finitedifference approx. to ∂ f/ ∂y	exact ∂f/ ∂ y
0.01
0.001
0.0001

Answer the followingquestions:

• For which partial derivative is the finite differenceapproximation more accurate?
• Why is the finite difference approximation for the otherpartial derivative less accurate?  Under what real-worldconditions might we see such poor approximations?