Symmetry of a line about the Origin

Question

I was reading about derivative of even and odd functions. I realized that If $f(x)$ be an odd function then $f'(x)$ is an even function. I'm trying to understand this intuitively. So I want to know that why the slope of the tangent line to function for the specific point $x_0>0$ is the same as slope of tangent line at $-x_0$ to the function. In fact I don't completely understand why symmetry of a line about the origin has the same slope with the original line.

Answer 1

For an odd function, $f(-\alpha)$ is obtained from $f(\alpha)$ by reflecting the point in the $y$-axis then in the $x$-axis, and vice-versa.

So, the tangent of an odd function at $(\alpha)$, first reflected in the $y$-axis then in the $x$-axis, gives its tangent at $(-\alpha),$ and vice-versa.

Since this process (in either direction) doesn't change the tangent's slope, we have that

$\quad$slope of odd function at $\alpha\;=\;$ slope of odd function at $(-\alpha),$

i.e., the derivative of an odd function is an even function.