Hello, mathaction!
I must assume you know the derivative of the arctangent.
. .  \;=\;\frac{u'}{1+u^2})
We have: . ![f(x)\;=\;[\arctan(3t)]^{-1}](http://latex.codecogs.com/png.latex?f(x)\;=\;[\arctan(3t)]^{-1})
Chain Rule: . ![f'(x)\;=\;-1[\arctan(3t)]^{-2}\cdot\frac{1}{1+(3t)^2}\cdot 3 \;=\;\frac{-3}{[\arctan(3t)]^2(1 + 9t^2)}<br />](http://latex.codecogs.com/png.latex?f'(x)\;=\;-1[\arctan(3t)]^{-2}\cdot\frac{1}{1+(3t)^2}\cdot 3 \;=\;\frac{-3}{[\arctan(3t)]^2(1 + 9t^2)}<br />
)
We have: . ![g(x)\;=\;\sin\left[\arcsin(\pi x)\right] \;=\;\pi x](http://latex.codecogs.com/png.latex?g(x)\;=\;\sin\left[\arcsin(\pi x)\right] \;=\;\pi x)
Therefore: .  \,=\,pi)
This one has a surprising answer . . .
We have: . \;=\;\frac{1}{1+x^2} + \frac{1}{1 + \left(\frac{1}{x}\right)^2}\cdot(-x^{-2}) \;= \;\frac{1}{1+x^2} + \frac{1}{1+\frac{1}{x^2}}\cdot\frac{-1}{x^2} )
. . 
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
How come it works out to zero?
Let )
Then: . 
is in a right triangle with: 
Here is the triangle: Code:
*
* *
* *
* * x
* *
* α *
* * * * * * * * * *
1
Let )
Then: . 
is in a right triangle with: 
Here is the triangle: Code:
*
* β*
* *
* * x
* *
* *
* * * * * * * * * *
1
Get it?
and
are in the same right triangle.
They are complementary: . 
We just differentiated a constant function: .  \:=\:\frac{\pi}{2})