components of vector valued functions

Given

F=@(x,y) [x+y,x-y.^2]

how do I refer to the second component of that function;

the command

syms x,y; F(x,y)(2)

gives an error, though I would expect to get x-y^2. My goal is to define a new function which only contains one particular component of a vector field like

f2=@(x,y) F(x,y)(2)

Probably simple but I did not find the correct syntax.

Answer 1

[Walter Roberson]

VSelect = @(V,IDX) V(IDX);
F = @(x,y) [x+y,x-y.^2]
f2 = @(x,y) VSelect(F(x,y),2);

There is no good direct way to do it. There is a way to do it all in one function call, but it is somewhat ugly and not easy to read.

[Wilhelm]

Thx! Not obvious, though ...

Answer 2

[Wilhelm]

That gives

>> syms z
>> F(1,z)

ans =

[ z + 1, 1 - z^2]

in my example.

I want to define a new function handle however which yields the second component x-y^2 only (for my example).

Or did I misunderstood your answer?