Find y' for y = arcsin x + arccos x
Find $y'$ for $y = \arcsin x + \arccos x$.
Explanation of Method
Two derivatives that cancel, revealing a hidden identity.
Step 1: Recall the standard derivatives
\[ (\arcsin x)' = \frac{1}{\sqrt{1-x^{2}}}, \qquad (\arccos x)' = -\frac{1}{\sqrt{1-x^{2}}}, \qquad |x| \lt 1. \]
The two differ only in sign, because $\arcsin$ increases exactly as fast as $\arccos$ decreases.
Step 2: Differentiate term by term
\[ y' = \frac{1}{\sqrt{1-x^{2}}} + \left(-\frac{1}{\sqrt{1-x^{2}}}\right) = 0. \]
Step 3: Verification
A function with zero derivative on an interval is constant there, so $y$ must be a constant — and evaluating at a convenient point identifies it: \[x = 0:\quad y = \arcsin 0 + \arccos 0 = 0 + \frac{\pi}{2} = \frac{\pi}{2}\]
\[x = \tfrac12:\quad y = \frac{\pi}{6} + \frac{\pi}{3} = \frac{\pi}{2} \;\;\checkmark\]
This recovers the classical identity
\[ \arcsin x + \arccos x = \frac{\pi}{2}, \qquad x \in [-1,\,1]. \]
Result
Final answer:
\[\boxed{\,y' = 0, \qquad |x| < 1\,}\]
Remark
Remark. The identity itself holds on the closed interval $[-1,1]$, but the derivative computation is valid only on the open interval $(-1,1)$: at the endpoints $x = \pm 1$ both derivatives become infinite, though their sum remains constant. This is a good illustration of differentiation as a tool for proving identities: show the derivative vanishes, then pin down the constant at one point.