Hello, mathaction!
I must assume you know the derivative of the arctangent.
. . $\displaystyle \frac{d}{dx}(\arctan u) \;=\;\frac{u'}{1+u^2}$
$\displaystyle f(x)\:=\:\frac{1}{\arctan(3t)}$
We have: .$\displaystyle f(x)\;=\;[\arctan(3t)]^{-1}$
Chain Rule: .$\displaystyle f'(x)\;=\;-1[\arctan(3t)]^{-2}\cdot\frac{1}{1+(3t)^2}\cdot 3 \;=\;\frac{-3}{[\arctan(3t)]^2(1 + 9t^2)}
$
$\displaystyle g(x)\;=\;\sin\left[\arcsin(\pi x)\right]$
We have: .$\displaystyle g(x)\;=\;\sin\left[\arcsin(\pi x)\right] \;=\;\pi x$
Therefore: .$\displaystyle g'(x) \,=\,pi$
$\displaystyle f(x) \;=\;\arctan(x) + \arctan\left(\frac{1}{x}\right)$
This one has a surprising answer . . .
We have: .$\displaystyle f'(x)\;=\;\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} $
. . $\displaystyle = \;\frac{1}{1+x^2} - \frac{1}{x^2+1} \;=\;0$
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
How come it works out to zero?
Let $\displaystyle \alpha = \arctan(x)$
Then: .$\displaystyle \tan\alpha \:=\:\frac{x}{1}\:=\:\frac{opp}{adj} $
$\displaystyle \alpha$ is in a right triangle with: $\displaystyle opp = x,\;adj = 1$
Here is the triangle: Code:
*
* *
* *
* * x
* *
* α *
* * * * * * * * * *
1
Let $\displaystyle \beta = \arctan\left(\frac{1}{x}\right)$
Then: .$\displaystyle \tan\beta \:=\: \frac{1}{x}\:=\:\frac{opp}{adj}$
$\displaystyle \beta$ is in a right triangle with: $\displaystyle opp = 1,\;adj = x$
Here is the triangle: Code:
*
* β*
* *
* * x
* *
* *
* * * * * * * * * *
1
Get it?
$\displaystyle \alpha$ and $\displaystyle \beta$ are in the same right triangle.
They are complementary: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$
We just differentiated a constant function: .$\displaystyle f(x) \:=\:\frac{\pi}{2}$