# Thread: Derivatives of Inverse Trigonometric Functions

1. ## Derivatives of Inverse Trigonometric Functions

I need to find the derivative of this problem, but am at a loss as to what to do.

h(t) = arccot(t) + arccot(1/t)

I know that h(t) = 0 but don't quite understand how to get this answer. If anyone could show me the steps I would greatly appreciate it!

2. Hello, NBrunk!

Do you know the derivative of the inverse cotangent?
If you do, exactly where is your difficulty?

Differentiate: . $h(t) \:=\: \text{arccot}(t) + \text{arccot}\left(\tfrac{1}{t}\right)$
. . $\boxed{\text{If }y \:=\:\text{arccot}(u),\,\text{ then: }\:\frac{dy}{dx} \:=\:\frac{\text{-}\,du}{1+u^2}}$

$\text{We have: }\;h(t) \;=\;\underbrace{\text{arccot}(t)}_{f(t)} + \underbrace{\text{arccot}\left(t^{-1}\right)}_{g(t)}$

. . and we want: . $h'(t) \;=\;f'(t) + g'(t)$

$f(t) \:=\:\text{arccot}(t)$

. . $f'(t) \;=\;\frac{-1}{1+t^2}$

$g(t) \;=\;\text{arccot}\left(t^{-1}\right)$

. . $g'(t) \;=\;\frac{-1}{1+t^{-2}}\cdot\left(-t^{-2}\right) \;=\;\frac{1}{1 + \frac{1}{t^2}}\cdot\frac{1}{t^2} \;=\;\frac{1}{t^2+1}$

Therefore: . $h'(t) \;=\;f'(t) + g'(t) \;=\;\frac{-1}{1+t^2} + \frac{1}{t^2+1} \;=\;0$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Here's an "eyeball" solution . . . which takes a bit of Thinking.

The first term is $\text{arccot}(t)$

This is an angle $\theta$ whose cotangent is $t.$

That is: . $\cot\theta \:=\:\frac{t}{1} \:=\:\frac{adj}{opp}$

Its right triangle looks like this:

Code:
                        * B
*  |
*     |
*        | 1
*           |
* θ            |
A * - - - - - - - - *
t

The second term is: . $\text{arccot}\left(\frac{1}{t}\right)$
This is an angle whose cotangent is: . $\frac{1}{t} \:=\:\frac{adj}{opp}$

Look at the diagram above . . . Can you find such an ngle?

With a little mental acrobatics, we see that it is angle $B$, the "other" angle!

You see: . $h(t) \;=\;\text{(an angle)} + (\text{its complement})$ . . . This sum is 90°, a constant.

So we did all that work to differentiate a constant!

3. Suppose $L=\tan(x)=\frac{\sin(x)}{\cos(x)}$, then $\tan\left(\tfrac{\pi}{2}-x\right)=\frac{\sin\left(\tfrac{\pi}{2}-x\right)}{\cos\left(\tfrac{\pi}{2}-x\right)}=\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x) }=\frac{1}{L}$. Therefore, $\tan\left(\tfrac{\pi}{2}-x\right)=\frac{1}{L}\implies \frac{\pi}{2}-x=\arctan\left(\tfrac{1}{L}\right)$, but $\tan(x)=L\implies x=\arctan\left(L\right)$ so we

may finally conclude that $\frac{\pi}{2}-\arctan\left(L\right)=\arctan\left(\tfrac{1}{L}\ri ght)$. Or, more pertinently $\frac{\pi}{2}=\arctan\left(L\right)+\arctan\left(\ tfrac{1}{L}\right)$.

So, as Soroban pointed our we have $\bigg(\arctan(x)+\arctan\left(\tfrac{1}{x}\right)\ bigg)\prime=\bigg(\tfrac{\pi}{2}\bigg)\prime=0$