The other four, from the quotient rule AB & BC

$\tan x = \dfrac{\sin x}{\cos x}$, so the quotient rule applies with $u = \sin x$ and $v = \cos x$, giving $u' = \cos x$ and $v' = -\sin x$:

$$\frac{d}{dx}\left[\tan x\right] = \frac{\cos x \cos x - \sin x(-\sin x)}{\cos^{2}x} = \frac{\cos^{2}x + \sin^{2}x}{\cos^{2}x} = \frac{1}{\cos^{2}x} = \sec^{2}x.$$

The step that decides the whole result is the middle one. The quotient rule subtracts, and $v' = -\sin x$ is itself negative, so the two minus signs combine into a plus. That is what produces $\cos^2 x + \sin^2 x$, which the Pythagorean identity collapses to 1.

Get that sign wrong and you get $\frac{\cos^2 x - \sin^2 x}{\cos^2 x}$, which simplifies to nothing useful and is not a standard result. An answer that refuses to simplify is a good signal to check the sign.

The result is $\sec^2 x$, which is positive wherever it is defined. That matches the graph: $\tan x$ is increasing on every branch.

$$\frac{d}{dx}\left[\cot x\right] = -\csc^{2}x, \qquad \frac{d}{dx}\left[\sec x\right] = \sec x\tan x, \qquad \frac{d}{dx}\left[\csc x\right] = -\csc x\cot x.$$

Each comes from the same rule. For the secant, $\sec x = \frac{1}{\cos x}$, so $u = 1$ and $v = \cos x$:

$$\frac{0 \cdot \cos x - 1(-\sin x)}{\cos^{2}x} = \frac{\sin x}{\cos^{2}x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x\tan x.$$

The split of $\frac{\sin x}{\cos^2 x}$ into the product $\sec x \tan x$ is the only non-mechanical step, and it is just regrouping.

For the cosecant the same work on $\frac{1}{\sin x}$ gives $\frac{-\cos x}{\sin^2 x} = -\csc x \cot x$. Here the numerator's derivative is 0 and $v' = \cos x$ is not negative, so nothing cancels the subtraction and the minus survives.

Lay the four out and one rule covers all the signs:

1. $\sin \to \cos$, and $\cos \to -\sin$.
2. $\tan \to \sec^{2}$, and $\cot \to -\csc^{2}$.
3. $\sec \to \sec\tan$, and $\csc \to -\csc\cot$.

The three functions whose names begin with co, cosine, cotangent and cosecant, are exactly the three whose derivatives are negative. The other three are positive. Nothing else needs remembering about the signs.

The second half of the pattern is that each derivative stays inside its own family. The tangent's derivative involves secants, and the cotangent's involves cosecants; the two families never mix. Pairing $\sec^2$ with $\cot$ is the single most common way this goes wrong.

Each result holds only where the function itself is defined: $\tan x$ and $\sec x$ fail at odd multiples of $\frac{\pi}{2}$, and $\cot x$ and $\csc x$ fail at multiples of $\pi$.

Once the four are known they behave like any other library derivative, so they combine with the sum, product and quotient rules directly.

$$\frac{d}{dx}\left[x\tan x\right] = \tan x + x\sec^{2}x,$$

by the product rule with $u = x$ and $v = \tan x$. Both terms survive, and the second one carries the $\sec^2 x$ intact.

$$\frac{d}{dx}\left[\frac{\sec x}{x}\right] = \frac{x\sec x\tan x - \sec x}{x^{2}},$$

by the quotient rule, where the leading term is the numerator's derivative, $\sec x \tan x$, multiplied by $x$.

One habit is worth keeping: when a question is stated in terms of $\tan$ or $\sec$, answer in those terms rather than converting everything back to sine and cosine. The identities are legal, but the graders and the answer keys both work in the stated family, and each conversion is another chance to lose a sign.