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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.

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  • $\begingroup$ You can also use related rates and model it physically. $\endgroup$
    – Sage Stark
    Jul 14, 2020 at 13:19

2 Answers 2

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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 $$(\dagger) = \frac{1}{\delta x}\tan^{-1}\frac{(x+ \delta x) + (-x)}{1 - (x+\delta x)(-x)}$$

$$ = \frac{1}{\delta x}\tan^{-1}\frac{\delta x}{1 + x^2 - x\delta x}$$

$$\rightarrow \frac{1}{1+x^2} \text{ by a small-angle approximation.}$$

enter image description here

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    $\begingroup$ I didn't know that Tan(a+b)=Tan(A)+Tan(B)/(1-Tan(A)Tan(B) now it makes sense I'm only a freshman in high school and haven't learned this knowledge before This proof is amazing. $\endgroup$
    – BriggyT
    Jul 13, 2020 at 18:04
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    $\begingroup$ Okay, I suppose this proof can be geometrical, because the tan formula can be proven using the corresponding formulae for sin and cos, which have geometrical proofs. en.m.wikipedia.org/wiki/… $\endgroup$ Jul 13, 2020 at 18:10
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    $\begingroup$ @Bladewood Done. How do you like it? $\endgroup$ Jul 14, 2020 at 2:37
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    $\begingroup$ @BenjaminWang It mostly looks good to me, but you want to avoid adding "EDIT" or "IN RESPONSE TO EDIT" in SE questions/answers because it feels more like a conversation thread than a Q/A. $\endgroup$
    – Bladewood
    Jul 14, 2020 at 14:31
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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} $

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    $\begingroup$ this makes sense I can follow this. $\endgroup$
    – BriggyT
    Jul 13, 2020 at 20:42

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