Does $A(x, m) = 1 - A(m, x)$ imply some symmetry in partial derivatives? This is a self-made problem. I have a function $A$ of variables $x, m \geq 0$. A satisfies the following symmetry:
$$A(x, m) = 1 - A(m, x)$$
Is there an equation that relates the partial derivative of $A$ wrt the first argument to the partial derivative of $A$ with respect to the second argument? It seems like there should be something straightforward but I'm not sure how to deduce it.
Thank you!
 A: All you need to do is differentiate both sides of the symmetry equation-
$$(\partial_1 A)(x,m) = \frac d{dx}[A(x,m)] = \frac d{dx}[1-A(m,x)] = - (\partial_2 A)(m,x)$$
or did you want something more?
A: I suggest using the definition of partial derivative to have a clearer point of view on the situation. The partial derivative of $A$ with respect to the first variable at the point $(x_0,m_0)$ is defined by :
$$\partial_1 A(x_0,m_0)=\lim_{h\to0} \frac{A(x_0+h,m_0)-A(x_0,m_0)}{h}.$$
Using the symmetry relation, we have :
$$\lim_{h\to0} \frac{(1-A(m_0,x_0+h))-(1-A(m_0,x_0))}{h}=\lim_{h\to0} \frac{-A(m_0,x_0+h)+A(m_0,x_0)}{h}.$$
The last term is equal, also by the definition partial derivative, to $- \partial_2 A(m_0,x_0)$.
Thus :
$$\partial_1 A(x_0,m_0)=- \partial_2 A(m_0,x_0)$$
A: Another way to look at this is to note that the function $B(x,m):=A(x,m)-\tfrac12$ is antisymmetric;
$$B(x,m)=A(x,m)-\tfrac12=(1-A(m,x))-\tfrac12=\tfrac12-A(m,x)=-B(m,x).$$
What does that tell you about the partial derivatives?
