Hello, Air!

Your thinking (and your notation) is a little fuzzy . . .

$\displaystyle f(x) \:= \:\sin^{-1}\left(\frac{x}{a}\right) \quad\Rightarrow\quad f'(x) \:= \:\frac{1}{\sqrt{a^2-x^2}}$

$\displaystyle f(x) \:= \:\cos^{-1}\left(\frac{x}{a}\right) \quad\Rightarrow\quad f'(x) \:= \:\frac{{\color{red}-}1}{\sqrt{a^2-x^2}}$

These two seem to imply that $\displaystyle \sin ^{-1} x + \cos ^{-1} x= 0$. . . . . no It says that the sum of their **derivatives** is zero.

$\displaystyle \text{Let: }\;\begin{array}{ccc}\alpha & = & \sin^{-1}\left(\frac{x}{a}\right) \\ \\[-3mm] \beta &=& \cos^{-1}\left(\frac{x}{a}\right) \end{array}$

We have this right triangle: Code:

*
* β|
a * |
* | x
* |
* α |
* - - - - - - - - *

We see that $\displaystyle \alpha$ and $\displaystyle \beta$ are in the *same right triangle.*

. . They are __complementary__: .$\displaystyle \alpha + \beta \:=\:\frac{\pi}{2}$

So: .$\displaystyle g(x) \;=\;\sin^{-1}\left(\frac{x}{a}\right) + \cos^{-1}\left(\frac{x}{a}\right) \;=\;\frac{\pi}{2}$

Now differntiate . . .

. . and you'll see why the sum of the *derivatives* is zero.