I want to calculate below matrix vector multiplication which involve differentiation.

$$\left( {\begin{array}{*{20}{c}} { - \mu \left( r \right)}&{\Delta {e^{ - i\phi }}\left( { - {\partial _r} + \frac{{i{\partial _\theta }}}{r}} \right)}\\ {\Delta {e^{i\phi }}\left( { - {\partial _r} + \frac{{i{\partial _\theta }}}{r}} \right)}&{\mu \left( r \right)} \end{array}} \right){e^{in\theta }}\left( {\begin{array}{*{20}{c}} {{e^{ - i\phi /2}}\left[ {f\left( r \right) + ig\left( r \right)} \right]}\\ {{e^{i\phi /2}}\left[ {f\left( r \right) - ig\left( r \right)} \right]} \end{array}} \right)$$

I use "basic math assitant" to write the following

enter image description here

However, it doesn't work. How to done it right?

PS. here you can download the notebook contains my expression

  • 1
    $\begingroup$ Please share the code in copyable form, so that other users can play with it. No one wants to retype all this code (and double check for correct transition). This will raise your chances for getting quick and competent help. $\endgroup$ Commented Jun 10, 2018 at 8:55
  • $\begingroup$ @MariuszIwaniuk Hi, The code directly copied is not perfect. So I attached a link to my notebook directly $\endgroup$
    – matheorem
    Commented Jun 10, 2018 at 9:03
  • 1
    $\begingroup$ we do not do links to notebooks at stackexchange. These could contains viruses. and when the link goes away, the question become useless. Code used should be posted in the question so it is self contained. $\endgroup$
    – Nasser
    Commented Jun 10, 2018 at 9:06

1 Answer 1


Just some toy example in order to give you an impression how this could be done.

A = Table[With[{t = t}, D[#, t] &], {t, {x, y}}, {i, 1, 2}];
U = {u[x, y], v[x, y]};
(Table[Sum[A[[i, j]]@U[[j]], {j, 1, 2}], {i, 1, 2}])//MatrixForm

$$\left( \begin{array}{c} u^{(1,0)}(x,y)+v^{(1,0)}(x,y) \\ u^{(0,1)}(x,y)+v^{(0,1)}(x,y) \\ \end{array} \right)$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.