The arc functions AB & BC

The six derivatives organize into three families, distinguished by what sits under the fraction:

1. Radical family. $\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}$, and $\arccos$ is its negative.
2. Rational family. $\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}$, and $\text{arccot}$ is its negative.
3. Absolute-value family. $\frac{d}{dx}\text{arcsec}\,x = \frac{1}{|x|\sqrt{x^2 - 1}}$, and $\text{arccsc}$ is its negative.

The rule for the mirrors is one sentence: every co-function derivative is the negative of its partner. That single fact converts three memorized forms into six.

The characteristic error is crossing the families, writing $\frac{d}{dx}\arctan x = \frac{1}{\sqrt{1-x^2}}$ or $\frac{d}{dx}\arcsin x = \frac{1}{1+x^2}$. A quick check settles it: $\arcsin$ is only defined for $|x|\le 1$, and its derivative should blow up as $x$ approaches $\pm 1$. The radical does that; $\frac{1}{1+x^2}$ does not. Meanwhile $\arctan$ is defined for all real $x$, so its derivative must never be undefined, and $1 + x^2$ is never zero.

Let $y = \arcsin x$, so $\sin y = x$ with $y$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. Differentiate implicitly:

$$\cos y\,\frac{dy}{dx} = 1 \quad\Longrightarrow\quad \frac{dy}{dx} = \frac{1}{\cos y}.$$

Now convert $\cos y$ into something written in $x$. Draw the right triangle where $\sin y = \frac{x}{1}$: the opposite side is $x$, the hypotenuse is 1, so the adjacent side is $\sqrt{1 - x^2}$. Therefore $\cos y = \sqrt{1 - x^2}$, and

$$\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}.$$

The positive root is the right one because cosine is non-negative on that restricted range. Running the same argument on $\tan y = x$ produces $\sec^2 y = 1 + x^2$ and gives the rational form. The two families differ because the Pythagorean identities behind them differ, which is a more durable memory than the shapes themselves.

These forms are stated for a bare $x$. When the argument is anything else, the chain rule applies exactly as it always does:

$$\frac{d}{dx}\arcsin(u) = \frac{u'}{\sqrt{1 - u^2}}, \qquad \frac{d}{dx}\arctan(u) = \frac{u'}{1 + u^2}.$$

So $\frac{d}{dx}\arctan(3x) = \frac{3}{1 + 9x^2}$. Two things move at once: the numerator gains the inner derivative 3, and the $x^2$ in the denominator becomes $(3x)^2 = 9x^2$. Changing only one of them is a common half-application. Squaring the argument means squaring all of it, so for $\arcsin(x^2)$ the denominator holds $1 - x^4$, not $1 - x^2$.

The derivative of $\arctan x$ is $\frac{1}{1+x^2}$, which is always positive and shrinks toward zero as $|x|$ grows. That matches the graph: $\arctan$ increases everywhere and flattens out toward its horizontal asymptotes.

The derivative of $\arcsin x$ is positive on $(-1, 1)$ and grows without bound at the ends, which matches a curve that rises steeply into its endpoints at $\pm\frac{\pi}{2}$. Reading a derivative back against the graph like this is the fastest way to catch a swapped form, and it is faster than re-deriving.