Find the value of a function whose derivative is zero The initial function is
$$h(x)=\arcsin  x + \arccos x$$
The derivative of this function is $0$
since $$h'(x)=\frac{1}{\sqrt{1-x^2}}-\frac{1}{\sqrt{1-x^2}}\equiv0$$  
This means that $h(x)$ is a constant function; how can I find the value of $h(x)$?   
Could anyone please explain?
 A: Try putting $x=0$ - since the function is constant, any convenient value will do.

Cautionary note: beware the domains of definition of the inverse trigonometric functions.
A: @MarkBennet gives the easiest (generally) way to figure out the value of a function once you know it's constant.
I'll only ellaborate here to give an intuitive reason for why this particular function is constant, which doesn't rely on any calculus:
$\arcsin x$ is the measure of the angle whose sine is $x$; $\arccos x$ is the measure of the angle whose cosine is $x$. Let's draw a picture to illustrate the first:

The measure of this angle is $\theta$, but more importantly, notice that this picture also includes $\arccos x$--the cosine of the other acute angle is also $x$.
But we know that the measure of the other angle is just $\frac{\pi}{2}-\theta$, so when we add the two measures together, we get
$$h(x)=\arcsin x + \arccos x=\theta+\left(\frac{\pi}{2}-\theta\right)=\frac{\pi}{2}$$
(basically, no matter what $\theta$ is, it cancels out)
(i.e., at core, this problem is just a fancy way of saying: in a right triangle, the two acute angles sum to $90^\circ$)
A: Note that $\cos(x)=\sin(\frac{\pi}{2}-x)$. Thus taking $x=\arccos(y)$ and taking the arcsine, we get $\arcsin(y)=\frac{\pi}{2}-\arccos(y)$, thus $$\arcsin(x)+\arccos(x)=\frac{\pi}{2}$$
