Function y-axis symmetry in (anti)derivatives If for a function f the equation f(x) = f( -x ) is true, is it also automatically true for all its derivatives and antiderivatives?
 A: The standard convention in English is that if $f(-x)=f(x)$ for all $x$, then $f$ is called an even function. If $f(-x)=-f(x)$ for all $x$, then $f$ is called an odd function.
A routine application of the Chain Rule shows that the derivative (if it exists) of an even function is odd, and that the derivative of an odd function is even.
But an antiderivative of an even function need not be odd, because of the constant of integration.
However, if $f$ is an even function that has an antiderivative, then $f$ has an odd antiderivative.
For suppose that $F'(x)=f(x)$ for all $x$, where $f$ is even.
Let 
$$G(x)=\frac{F(x)-F(-x)}{2}.$$
It is easy to see that $G$ is odd. But note that
$$G'(x)=\frac{1}{2}\left(f(x)-(-1)f(-x)\right)=f(x).$$
A: For the derivatives of $f$: use the chain rule. Symmetric functions turn into antisymmetric functions and vice-versa under differentiation. We can use this to generalize to higher-order derivatives.
If $F$ is an antiderivative of a nonzero function $f$, then it cannot be symmetric (or else its derivative, which is $f$, would be antisymmetric, but it isn't). Is it necessarily antisymmetric? No, because if $F$ is antisymmetric and nonzero then $F+C$ for a nonzero constant $C$ is not antisymmetric (we would have $-\big(F(x)+C\big)=-F(x)+C$, which implies $C=0$), yet $F+C$ is yet another antiderivative of $f$.
However, there can be a particular antisymmetric antiderivative, as $F=\sin$ and $f=\cos$ exhibits.
