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.


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.

