Hello, NBrunk!
Do you know the derivative of the inverse cotangent?
If you do, exactly where is your difficulty?
Differentiate: .  \:=\: \text{arccot}(t) + \text{arccot}\left(\tfrac{1}{t}\right))
. . ,\,\text{ then: }\:\frac{dy}{dx} \:=\:\frac{\text{-}\,du}{1+u^2}})
 \;=\;\underbrace{\text{arccot}(t)}_{f(t)} + \underbrace{\text{arccot}\left(t^{-1}\right)}_{g(t)})
. . and we want: .  \;=\;f'(t) + g'(t))
 \:=\:\text{arccot}(t))
. .  \;=\;\frac{-1}{1+t^2})
 \;=\;\text{arccot}\left(t^{-1}\right) )
. .  \;=\;\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: .  \;=\;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 )
This is an angle 
whose cotangent is 
That is: . 
Its right triangle looks like this:
Code:
* B
* |
* |
* | 1
* |
* θ |
A * - - - - - - - - *
t
The second term is: . )
This is an angle whose cotangent is: . 
Look at the diagram above . . . Can you find such an ngle?
With a little mental acrobatics, we see that it is angle 
, the "other" angle!
You see: .  \;=\;\text{(an angle)} + (\text{its complement}))
. . . This sum is 90°, a constant.
So we did all that work to differentiate a constant!
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