The four library derivatives AB & BC

$$\frac{d}{dx}\left[\sin x\right] = \cos x, \qquad \frac{d}{dx}\left[\cos x\right] = -\sin x.$$

The minus sign sits on the cosine's derivative, not the sine's. The picture confirms it: at $x = 0$ the graph of $\sin x$ is climbing at its steepest, and $\cos 0 = 1$; at that same input the graph of $\cos x$ is at a maximum and momentarily flat, and $-\sin 0 = 0$.

A one-second check for the sign, available in any exam: just to the right of $x = 0$, $\cos x$ is decreasing, so its derivative there must be negative. $-\sin x$ is negative for small positive $x$; $+\sin x$ would be positive and would say the cosine is rising, which the graph flatly contradicts.

Both statements require $x$ in radians. In degrees an extra factor of $\frac{\pi}{180}$ appears, which is the reason radians are the default everywhere in calculus.

The derivative of $\sin x$ comes straight from the difference quotient, using the angle addition formula and the two limits established in Topic 1.7:

$$\frac{\sin(x + h) - \sin x}{h} = \cos x \cdot \frac{\sin h}{h} + \sin x \cdot \frac{\cos h - 1}{h} \longrightarrow \cos x \cdot 1 + \sin x \cdot 0 = \cos x.$$

Differentiating repeatedly runs a four-step cycle and returns to the start:

$$\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x.$$

So $\frac{d^2}{dx^2}[\sin x] = -\sin x$, and the fourth derivative of $\sin x$ is $\sin x$ again. The minus signs enter at the second and third steps and leave at the fourth. Losing track of where you are in the cycle is the usual cause of a sign error in a higher derivative.

$$\frac{d}{dx}\left[e^{x}\right] = e^{x}, \qquad \frac{d}{dx}\left[a^{x}\right] = a^{x}\ln a.$$

$e^x$ is the unique exponential function that is its own derivative, and that property is what makes $e$ worth defining. Every other base carries a factor of $\ln a$: for $a = 2$ that factor is about $0.693$, and for $a = 10$ it is about $2.303$.

Two boundaries are worth restating from Topic 2.5. The power rule does not touch either of these, because the variable is the exponent. And the constant multiple rule does apply, so $\frac{d}{dx}[5e^x] = 5e^x$.

Note also that $e^x$ is never zero and always positive, so its derivative is too: the graph is increasing everywhere, which no sign error should ever contradict.

$$\frac{d}{dx}\left[\ln x\right] = \frac{1}{x} \ \ (x > 0), \qquad \frac{d}{dx}\left[\log_{b} x\right] = \frac{1}{x \ln b}.$$

Only the natural log gives the clean $\frac{1}{x}$. Any other base picks up $\ln b$ in the denominator, which is the mirror image of what the exponential rule does with it. The two are related by the change of base identity $\log_b x = \frac{\ln x}{\ln b}$, where $\frac{1}{\ln b}$ is just a constant multiple.

Three details that get lost:

1. The domain. $\ln x$ exists only for $x > 0$, so $\frac{1}{x}$ is its derivative on that domain and the question of $x < 0$ does not arise.
2. The result is a power, not a log. Differentiating a logarithm produces a rational function; nothing logarithmic survives in the answer.
3. $\ln$ is a function, not a multiplier. $\ln(5x)$ is not $5 \ln x$, although a log property does turn it into $\ln 5 + \ln x$, and that rewrite makes its derivative $\frac{1}{x}$ by inspection.