Derivatives of Inverse Trig Functions

Let $y = \arcsin x$, so $\sin y = x$. For $|x| < 1$, $-\frac{\pi}{2} < y < \frac{\pi}{2}$, where $\cos y > 0$. Differentiating implicitly, $\cos y \cdot \frac{dy}{dx} = 1$, so $\frac{dy}{dx} = \frac{1}{\cos y}$, and $\cos y = \sqrt{1 - \sin^2 y} = \sqrt{1 - x^2}$, as the left triangle shows. This is the inverse-function rule $\left(f^{-1}\right)'(x) = \frac{1}{f'\left(f^{-1}(x)\right)}$ with $f = \sin$ on $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$, where $f'(y) = \cos y \ne 0$. That rule is also what guarantees $\arcsin$ is differentiable for $|x| < 1$.

For $y = \arctan x$, $\tan y = x$ with $-\frac{\pi}{2} < y < \frac{\pi}{2}$. Then $\sec^2 y \cdot \frac{dy}{dx} = 1$, and $\sec^2 y = 1 + \tan^2 y = 1 + x^2$, which the right triangle shows. So $\frac{dy}{dx} = \frac{1}{1 + x^2}$.

Arccos and arccot follow from identities. The number $\frac{\pi}{2} - \arcsin x$ lies in $[0, \pi]$, and its cosine is $\sin(\arcsin x) = x$, so it equals $\arccos x$. In the same way, $\operatorname{arccot} x = \frac{\pi}{2} - \arctan x$. So their derivatives are the negatives of those above.

For $|x| > 1$, let $y = \operatorname{arcsec} x$, so $\sec y = x$, with $y$ in $\left(0, \frac{\pi}{2}\right)$ or $\left(\frac{\pi}{2}, \pi\right)$. Differentiating implicitly, $\sec y \tan y \cdot \frac{dy}{dx} = 1$. On the first interval, $\sec y$ and $\tan y$ are both positive, and on the second they're both negative. Either way their product is positive, so $\sec y \tan y = |\sec y|\,|\tan y|$. Since $\tan^2 y = \sec^2 y - 1 = x^2 - 1$, that product is $|x|\sqrt{x^2 - 1}$. So $\frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2 - 1}}$. The cosecant works the same way.