Symmetry of a line about the Origin 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.

Write $f(x)$ is a Taylor polynomial around $0$. Then if $f(x)$ is an odd function, the highest-order term has an odd power, say $x^{2k+1}$, $k \in \mathbb Z^+$, as $(-x)^{2k+1} = -x^{2k+1}$, which satisfies the definition for an odd function.
Thus the highest-order term of $f'(x)$ is $\frac{d}{dx} x^{2k  +1} = x^{2k}$, and $f'(x)$ is an even function. You can check this by verifying if $f'(-x) = f'(x)$.
Note that this only applies around the radius of convergence (if $f(x)$ is not a polynomial).
