Aakanchha D.

asked • 06/10/15

I need to get the value of this combined partial derivatives in term of individual partial derivatives

How can i write d/dy(d/dx(f(x,y))*delta x*delta y in terms of        d/dx(f(x,y))*delta x    and d/dy(f(x,y))*delta y.
Even an approx value will work

Aakanchha D.

see, The issue is that if I have a software that takes in value of:1) f(x,y); 2) d/dx(f(x,y))*delta x; and 3)d/dy(f(x,y))*delta y independently then how can i get that mixed partial differential values in terms of these three  values only?
 
suppose if my f(x,y)= xy   say it to be C;
then my d/dx(f(x,y))=y say it to be A; d/dy(f(x,y))=x say it to be B; d/dx(d/dy(f(x,y)))=1 say it to be D; then I can write this mixed differential as D= A*B/C;
 
I need a genralise formula for any such function
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06/10/15

Keith M.

As long as you are guaranteed that the mixed partial exists at a point, you can simply use one or the other first partial derivative (A or B) to approximate the mixed partial at that point using the symmetric quotient difference method.
 
The gist of the algorithm is to use the approximate definition of the derivative:
2*Δ*∂xi(f(x1,x2,...,xi,...)) = f(x1,x2,...,xi+Δ,...)-f(x1,x2,...,xi-Δ,...)
which converges to the value of the derivative in the limit as Δ approaches 0 (assuming the derivative exists at that point).
 
Since we can't actually evaluate that limit, we just try smaller and smaller values of Δ until the value of ∂xi begins to converge (e.g. when we halve Δ, ∂xi changes by no more than .001, although you can try out different threshold values here).  The one thing to watch out for is that at some points the mixed partial may be undefined!  In this case, your algorithm may need to set a cap on the number of times Δ can be halved (e.g. 1000 iterations) or a cap on the value of ∂xi (e.g. |∂xi| < 1,000,000).  Note, though, that these restrictions are arbitrary and do not guarantee correctness for all inputs.
 
This is the same algorithmic approach taken by most mid-range graphing calculators (e.g. TI-83) to approximate derivatives of functions with respect to one variable.  Although it works well in most cases, it doesn't work everywhere and will get fooled by spots where the derivative does not exist (e.g. TI-83's nDeriv function evaluates to 0 for ∂x(3√x2)|x=0).
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06/11/15

1 Expert Answer

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Keith M. answered • 06/10/15

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This problem is a straightforward review of partial derivatives.  There is a nice guide with relevant definitions and examples on WolframAlpha for those interested (http://mathworld.wolfram.com/PartialDerivative.html).
 
I will use ∂x(f(x,y)) to denote the partial derivative of f with respect to x, and ∂y(f(x,y)) to denote the partial derivative of f with respect to y.  Similarly, ∂xy(f(x,y)) represents the (second order) mixed partial derivative that you are trying to represent in terms of the two (first order) partials.
 
The link above provides your answer: for continuous functions f(x,y) whose first and second order partials are also continuous, ∂xy(f) = ∂x(∂y(f)) = ∂y(∂x(f)).
 
Note that in general, though, for functions f(x,y) which are discontinuous at (x,y) or functions f(x,y) whose first and second order partials are discontinuous at (x,y), ∂xy(f) = ∂x(∂y(f)) ≠ ∂yx(f) = ∂y(∂x(f)), and there is no way to write ∂xy(f) in terms of ∂x(f).  Like the one-dimensional cases you've studied in introductory calculus, though, polynomial functions f(x,y) will always be continuous and differentiable, meaning it doesn't matter which order you take the partials in-- you'll get the same result.
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