# Derivatives of Inverse Trigonometric Functions

• Jan 14th 2010, 06:01 PM
NBrunk
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!
• Jan 14th 2010, 07:01 PM
Soroban
Hello, NBrunk!

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

Quote:

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

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

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

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

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

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

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

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

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

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

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

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

Its right triangle looks like this:

Code:

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

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

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

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

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

So we did all that work to differentiate a constant!

• Jan 14th 2010, 07:43 PM
Drexel28
Suppose $\displaystyle L=\tan(x)=\frac{\sin(x)}{\cos(x)}$, then $\displaystyle \tan\left(\tfrac{\pi}{2}-x\right)=\frac{\sin\left(\tfrac{\pi}{2}-x\right)}{\cos\left(\tfrac{\pi}{2}-x\right)}=\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x) }=\frac{1}{L}$. Therefore, $\displaystyle \tan\left(\tfrac{\pi}{2}-x\right)=\frac{1}{L}\implies \frac{\pi}{2}-x=\arctan\left(\tfrac{1}{L}\right)$, but $\displaystyle \tan(x)=L\implies x=\arctan\left(L\right)$ so we

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

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