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.
 A: 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.
A: 
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.


The reason for this thing lies in the fact that even function are symmetric with respect to y axis, that's why $f'(x)=f'(-x)$, the slope of the tangent is same at both the sides.
For better understanding assume that $f(x)=\sin x$ therefore $f'(x)=\cos x$
Now lets draw tangent to two points on $\sin x$ :- $(\pi/4,1/\sqrt2)$, $(-\pi/4,-1/\sqrt2)$
Now the slope of tangent will be same for both the points only because graph of $\cos x$ is symmetric with respect to y axis so value of slope i.e. value of $\cos x$ will be same for both $\pi/4$ and $-\pi/4$. Here is a graph for you to understand better.

A: 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).
