How do you prove that the derivative $\tan^{-1}(x)$ is equal to $\frac{1}{1+x^2}$ geometrically How do you prove that the derivative of $\tan^{-1}(x)$ is equal to $\frac{1}{1+x^2}$ geometrically?
I figured it out by working it out using implicit differentiation.
I also found how to plot a semi-circle using $\cos^2(x)+\sin^2(x)=1$ and found that $((x^2-1))^{0.5}$ plots a semi-circle because if you wanted to find $\sin(x)$ with $\cos(x)$ you would do $(\cos(x)^2-1)^{0.5}$. The reason only a semi circle is in order for it to work it must be both the positive and negative solutions.
I saw that $1$ and the $x^2$ and thought you could visually see that the derivative of $\tan^{-1}(x)$ is $\frac{1}{1+x^2}$ but I couldn't find any way so far.
 A: Another proof comes from
the atan addition/subtraction formula
$\arctan(a)\pm\arctan(b)
=\arctan(\frac{a\pm b}{1\mp ab}
$.
Then,
with a little hand-waving
as $h \to 0$
and assuming that
$\lim_{z \to 0} \dfrac{\arctan(z)}{z}
=1
$,
$\begin{array}\\
\arctan(x+h)-\arctan(x)
&=\arctan(\dfrac{x+h-x}{1+x(x+h)})\\
&=\arctan(\dfrac{h}{1+x(x+h)})\\
&\to\arctan(\dfrac{h}{1+x^2})\\
\text{so}\\
\dfrac{\arctan(x+h)-\arctan(x)}{h}
&\to\dfrac{\arctan(\dfrac{h}{1+x^2})}{h}\\
&=\dfrac1{1+x^2}\dfrac{\arctan(\dfrac{h}{1+x^2})}{\frac{h}{1+x^2}}\\
&\to\dfrac1{1+x^2}\\
\end{array}
$
A: Tried to do geometry, but ended up doing a hand-wave-y first-principles approach. It can be made rigorous though.
The angle addition formulae have geometrical proofs.
$$\text{We want to find }\ \ \ (\dagger) :=  \frac{1}{\delta x} (\tan ^{-1}(x + \delta x) - \tan ^{-1} x) \ \ \ \text{ as } \delta x \rightarrow 0.$$
Recall that $$\tan(A+B)= \frac{\tan A + \tan B}{1-\tan A \tan B} \ ,$$
we can substitute $u = \tan A$ and $v = \tan B$ to get
$$\tan^{-1}u + \tan^{-1} v = \tan^{-1} \frac{u+v}{1-uv} \ .$$
Therefore $$(\dagger) = \frac{1}{\delta x}\tan^{-1}\frac{(x+ \delta x) + (-x)}{1 - (x+\delta x)(-x)}$$
$$ = \frac{1}{\delta x}\tan^{-1}\frac{\delta x}{1 + x^2 - x\delta x}$$
$$\rightarrow \frac{1}{1+x^2} \text{ by a small-angle approximation.}$$

