The arctangent function:

The arctangent function, $\arctan$, is the inverse of the Principle Tangent Function, ${\rm Tan}$, defined above. Algebraically this means $\arctan({\rm Tan}(a)) = a$ and $\rm{Tan}(\arctan(a)) = a$ so long as domain requirements are respected. To find the slope of the tangent line to $y = \arctan(x)$ at the point $(a,\arctan(a))$ we reason that if the slope of this line is m, then the slope of the tangent line to $y = {\rm Tan}(x)$ at $(\arctan(a),a)$should be 1/m, since when the graph of $y = {\rm Tan}(x)$ is reflected onto the graph of $y = \arctan(x)$, the point $(\arctan(a),a)$ is carried onto the point $(a,\arctan(a))$. Well, since ${\rm Tan}'(x) = \sec^2(x)$, we know that the slope of the tangent line to $y = {\rm Tan}(x)$ at $(\arctan(a),a)$ is given by $\sec^2(\arctan(a))$. Surprisingly, this expression can be simplied. If you look at this picture of a right triangle,

you see that if $\theta = \arctan(a)$ then the side opposite $\theta$ can have length a and the side adjacent to $\theta$ can have length 1. This forces the hypoteneuse to have length $\sqrt{1+a^2}$. Now we see that the secant of $\theta$ is $\sqrt{1+a^2}/1$, so $\sec^2(\arctan(a)) = 1 + a^2$. Hence 1/m = (1+a²), that is m = 1/(1+a²) is the slope of the tangent line to $y = \arctan(x)$ at $(a,\arctan(a))$. Hence we find that the rate of change of $\arctan$ is given by $\arctan'(a) = 1/(1+a^2)$!