For every fuction with symmetry will the first derivative have symmetry of the other type? What Theorem proves this if it is in fact true?Yes, it is true. If f is an even function (that is, has the same value if you replace x by -x), then its derivative will be an odd function (changes sign when you replace x by -x), and vice versa.
This is quite clear geometrically; in the picture below, for example, it is apparent that the slopes m and M are negatives of each other. You could even turn this into a geometric proof: if f is even, its graph is the same if you reflect it in a mirror placed along the y-axis, and therefore the tangent line at one point is the mirror reflection of the tangent line at the reflected point, and a line reflected in the y-axis has its slope multiplied by -1.
* | *
slope m -->\* | */<-- slope M
* | *
***
|
------------+----------------
-x x
In the above picture, m = -M.
The way that you prove it using only calculus theorems (without needing any geometry at all) is as follows.
If f is an even function, that means that f(x) = f(-x).
Now differentiate both sides. The left-hand side becomes f'(x), and the right-hand side becomes -f'(-x) (using the chain rule).
Therefore, f'(x) = - f'(-x). In other words, the value of f' at x is the negative of its value at -x, so f' is an odd function.
Similarly, if you started with an odd function f, you have f(x) = - f(-x). Differentiating both sides gives f'(x) = + f'(-x), so f' is an even function.
In either case, f' has the opposite type of symmetry (even or odd) from f.