Symbol matrix function differentiation

2013 3 3
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1.The function 'diff' seems could only work with one-element variable 'v',
diff(f,v); % f(v), v is a one-lelement variable
while I would like do differentiation on a symbol matrix
diff(F, V); % F(V), V is a symbol matrix variable
2.If I make differentiation to each element of the symbol matrix, and obtain the result of diff(F, V), while the result is in element by element format,
[g1(v_i), g2(v_i), ..., gn(v_i)]
so I want to know are there some methods to make the result in symbol matrix variable format like this?
g(V)
3.for example
diff(x^T*A*x, x) = A^T*x + A*x; % A is a constant matrix, x is a vector

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Walter Roberson
Walter Roberson 2013-3-4
编辑：Walter Roberson 2013-3-4

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If you want the result to be like in your (2) above, then use a "for" loop, or use
g = feval(symengine, 'zip', F, V, 'diff');
or use
g = cell2mat(arrayfun(@diff, F, V, 'Uniform', 0));
You might be able to just use
g = arrayfun(@diff, F, V);
but I think I recall seeing that have problems with arrayfun expecting numeric results.

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Youssef Khmou
Youssef Khmou 2013-3-4
编辑：Youssef Khmou 2013-3-4
hi Walter, the first & last options :
g=arrayfun(@diff, F, 2) % or V generally
g = cell2mat(arrayfun(@diff, f, 1, 'Uniform', 0))
do not work :
??? Error using ==> arrayfun
Unsupported ARRAYFUN input type: sym
Walter Roberson
Walter Roberson 2013-3-4
Ah. Well there are always the syntax-disguised for loops:
g = reshape( cell2mat( arrayfun(@(K) diff(F(K), V(K)), 1:numel(F), 'Uniform', 0) ), size(F))
chen xy
chen xy 2013-3-4
Hi Walter, thx for answering. I already could obtain the result liked in (2) by using 'for' loop, what I want to do is like this: 1.type the command
diff(x.'*A*x, x)
2.the output is better as the follow
ans =
A.'*x + A*x
It seems be a very difficult work for implementation, I'd better deduce the formula on the paper as assistance.
Walter Roberson
Walter Roberson 2013-3-4
If the .' represents transpose, then how could that be the correct output if A is not square?
chen xy
chen xy 2013-3-5
If A isn't square, "x.'*A*x" will be wrong for matrix dimensions.

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