Derivatives of cos x, sin x, eˣ, and ln x

The power rule handles $x^n$, where the base varies and the exponent is fixed. It says nothing about $\sin x$, about $e^x$ (where the exponent varies), or about $\ln x$. Each of these four functions gets its derivative from the limit definition and one special limit, and the results are short: the derivative of $\sin x$ is $\cos x$, of $\cos x$ is $-\sin x$, of $e^x$ is $e^x$ itself, and of $\ln x$ is $\frac{1}{x}$.

Sine and cosine

Start with the picture. The graph of $y = \sin x$ below has slope 1 as it crosses the origin, slope 0 at its peak at $x = \frac{\pi}{2}$, slope $-1$ crossing $x = \pi$, and slope 0 again at its low point at $x = \frac{3\pi}{2}$. Those slopes, $1, 0, -1, 0, 1$, are exactly the values of $\cos x$ at the same points.

To prove it, use the angle-sum identity $\sin(x+h) = \sin x\cos h + \cos x\sin h$ in the difference quotient and group the terms by what multiplies them:

As $h \to 0$, $\sin x$ and $\cos x$ don't change, because they don't involve $h$. Two limits remain. The first, $\lim\limits_{h\to 0}\frac{\sin h}{h} = 1$, comes from the squeeze theorem (topic 1.8). The second follows from it: multiplying top and bottom by $\cos h + 1$ and using $\cos^2 h - 1 = -\sin^2 h$,

So the difference quotient approaches $\sin x\cdot 0 + \cos x\cdot 1 = \cos x$. The same steps with $\cos(x+h) = \cos x\cos h - \sin x\sin h$ give $\cos x\cdot 0 - \sin x\cdot 1 = -\sin x$.

The minus sign on the cosine derivative matches the graph: on $(0, \pi)$, where $\sin x > 0$, the cosine curve is falling, so its slope $-\sin x$ is negative there.

These formulas need $x$ in radians. The limit $\frac{\sin h}{h} \to 1$ is true only when $h$ is measured in radians; with degrees the limit is $\frac{\pi}{180}$, and every derivative would carry that factor. Calculus assumes radians throughout, so a calculator used for calculus must be in radian mode.

The exponential function $e^x$

For any base $b > 0$, the law of exponents $b^{x+h} = b^x b^h$ lets $b^x$ factor out of the difference quotient:

The remaining fraction doesn't involve $x$, so its limit is a single number for each base; call it $L(b)$. Then $\frac{d}{dx}b^x = L(b)\,b^x$: every exponential function grows at a rate proportional to its own height. Small values of $h$ estimate $L(b)$. For $b = 2$ the fraction is about $0.693$, and for $b = 3$ it is about $1.099$. One base between 2 and 3 makes the constant exactly 1, and that base is $e \approx 2.71828$. So

The graph of $y = e^x$ therefore has slope equal to its height at every point: slope 1 at $(0, 1)$, slope $e^2 \approx 7.389$ at $(2, e^2)$. Since $e^x > 0$ everywhere, the slope is always positive and the function is always increasing. (Other bases have $L(b) = \ln b$, so $\frac{d}{dx}b^x = b^x\ln b$; topic 3.1 shows why.)

Compare $e^x$ with $x^e$. In $x^e$ the exponent is the constant $e$, so the power rule applies and the derivative is $e\,x^{e-1}$. In $e^x$ the variable is in the exponent, and the power rule doesn't apply at all.

The natural logarithm $\ln x$

The natural logarithm $\ln x$ is the logarithm with base $e$: $\ln x = y$ means $e^y = x$. It is defined only for $x > 0$, and its graph is the graph of $y = e^x$ reflected across the line $y = x$.

That reflection gives the derivative. Take any point $(a, e^a)$ on the graph of $e^x$; its tangent line has slope $e^a$, meaning a rise of $e^a$ for every run of 1. Reflecting across $y = x$ swaps every horizontal distance with a vertical one, so the reflected tangent line, which touches the graph of $\ln x$ at $(e^a, a)$, has a rise of 1 for every run of $e^a$, a slope of $\frac{1}{e^a}$. At that point the $x$-coordinate is $e^a$, so the slope there is $\frac{1}{x}$. Topic 3.3 turns this picture into the general rule for derivatives of inverse functions.

The derivative $\frac{1}{x}$ is positive for every $x > 0$, so $\ln x$ is always increasing, and it shrinks toward 0 as $x$ grows, so the graph keeps rising but flattens out. Near $x = 0$ the slope is huge, matching the graph's steep drop toward its vertical asymptote.

Logarithm properties combine with these rules to handle more forms before you learn the chain rule. Because $\ln(5x) = \ln 5 + \ln x$, its derivative is $0 + \frac{1}{x} = \frac{1}{x}$: scaling the input of $\ln$ by 5 shifts its graph up by $\ln 5$ and changes no slope. Because $\ln\left(x^3\right) = 3\ln x$ for $x > 0$, its derivative is $\frac{3}{x}$.

Limits that are derivatives in disguise

The definition of the derivative at $x = a$ comes in two forms:

Read backward, this means some limits are derivatives you already know. Every such limit has the indeterminate form $\frac{0}{0}$, so plugging in fails, but if you recognize the shape you can evaluate it with a derivative rule. To match a limit to the definition:

1. Find the function $f$ and the point $a$: the expression being evaluated at $a + h$ (or at $x$) tells you $f$, and the point it approaches tells you $a$.
2. Check that the term being subtracted really is $f(a)$.
3. The limit equals $f'(a)$. Differentiate $f$ and evaluate at $a$.

For example, $\lim\limits_{h\to 0}\frac{e^{2+h} - e^2}{h}$ has $f(x) = e^x$ and $a = 2$, and the subtracted term $e^2$ is $f(2)$, so the limit is $f'(2) = e^2$. And the limit $\lim\limits_{x \to 0}\frac{\sin x}{x} = 1$ from topic 1.8 is itself one of these: since $\sin 0 = 0$, it is the derivative of $\sin x$ at 0, which is $\cos 0 = 1$.

Worked examples

Where students lose points

On the exam

These four derivatives are expected from memory on both sections; no formula sheet gives them. A limit written as a difference quotient asks for $f'(a)$; recognizing the function and the point is the whole task, and no work beyond the derivative rule is needed. On calculator-active questions, the calculator can evaluate a derivative at a point, but any equation like $\frac{1}{x} = e^x$ should be written down before you solve it.