역삼각함수의 미분

Using formulas you should memorize from this section, $$\frac{\mathrm{d}}{\mathrm{d}x}\{\arcsin x + \arccos x\}=\frac{1}{\sqrt{1-x^2}}+\frac{-1}{\sqrt{1-x^2}}=0$$

Remark: the only functions with derivative equal to zero everywhere are constant functions, so $\arcsin x + \arccos x$ should be a constant. Since $\sin \theta = \cos \left(\frac{\pi}{2}-\theta\right)$, we can set

$$\begin{aligned} \sin\theta&=x & \cos\left(\frac{\pi}{2}-\theta\right)&=x\\ \text{where $x$ and $\theta$ are the same in both expressions, and $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$. Then} \\ \arcsin x &=\theta & \arccos x &= \frac{\pi}{2}-\theta \\ \text{We note here that arcsine is the inverse of the sine function \textit{ restricted to} $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. So, since we restricted $\theta$ to this domain, $\sin \theta=x$ really does imply $\arcsin x = \theta$. (For an example of why this matters, note $\sin(2\pi)=0$, but $\arcsin (0)=0 \neq 2\pi$.) Similarly, arccosine is the inverse of the cosine function restricted to $[0,\pi]$. Since $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$, then $0 \leq (\frac{\pi}{2}-\theta) \leq \pi$, so $\cos\left(\frac{\pi}{2}- \theta\right) =x$ really does imply $\arccos x=\frac{\pi}{2}-\theta$.} \\ \end{aligned}$$

So, $$\arcsin x+\arccos x =\theta+\frac{\pi}{2}-\theta =\frac{\pi}{2}$$ which means the derivative we were calculating was actually just $\displaystyle \frac{\mathrm{d}}{\mathrm{d}x}\left\{\dfrac{\pi}{2}\right\}=0$.