Derivatives of sin x, cos x, e to… — AP® Calculus AB Notes

Polynomials are only the beginning. Waves, growth and decay are described by sine, cosine, the exponential $e^x$ and the natural logarithm $\ln x$ . Their derivatives are short and worth memorizing, and each one makes sense from a picture. This page gives all four, shows where they come from, and uses them to evaluate limits that are derivatives in disguise.

Sine and cosine

Look at the graph of $y = \sin x$ and track its slope. At $x = 0$ it rises with slope 1. At $\frac{\pi}{2}$ , the top of the wave, its tangent line is level: slope 0. At $\pi$ it falls with slope $-1$ . At $\frac{3\pi}{2}$ , the bottom, the slope is 0 again, and at $2\pi$ it is back to 1. The list $1, 0, -1, 0, 1$ is exactly the list of heights of $\cos x$ at those inputs.

Top: y = sin x from 0 to 2 pi with short tangent segments at 0, pi/2, pi, 3pi/2 and 2pi, whose slopes are 1, 0, minus 1, 0 and 1. Bottom, on the same scale: y = cos x, whose heights at those x values are 1, 0, minus 1, 0 and 1. The derivative of sin x is cos x. — **Figure 1.** Each slope of $\sin x$ (top) is the height of $\cos x$ (bottom) at the same $x$ . LevlPrep original diagram.

That match holds at every $x$ :

$\frac{d}{dx}(\sin x) = \cos x, \qquad \frac{d}{dx}(\cos x) = -\sin x.$

The minus sign for cosine is easy to check: just right of $x = 0$ , the graph of $\cos x$ starts to fall, so its slope is negative, while $\sin x$ is positive there.

Why is the slope of $\sin x$ at 0 exactly 1? The difference quotient at 0 is $\frac{\sin h - \sin 0}{h}$ $= \frac{\sin h}{h}$ , and Unit 1 showed with the squeeze theorem that $\lim_{h \to 0} \frac{\sin h}{h}$ $= 1$ . That limit is true only when angles are measured in radians, so these formulas are radian formulas. In degrees, every slope would pick up an extra factor of $\frac{\pi}{180}$ .

The exponential $e^x$

Every exponential graph $b^x$ has a slope equal to a fixed multiple of its height. For one special base, the slope at $x = 0$ is exactly 1, and then the slope at every point equals the height. That base is $e \approx 2.718$ , and so

$\frac{d}{dx}\left( e^x \right) = e^x.$

The function is its own derivative. At $x = 0$ the slope is 1; at $x = \ln 3$ the height is 3 and so is the slope. Do not use the power rule here: in $e^x$ the variable is the exponent, so $xe^{x - 1}$ is wrong.

The natural logarithm

For $x > 0$ ,

$\frac{d}{dx}(\ln x) = \frac{1}{x}.$

The graph of $\ln x$ is steep near 0 and flattens as $x$ grows, matching $\frac{1}{x}$ : large for small $x$ , small for large $x$ , and always positive, because $\ln x$ always increases. Unit 3 shows why this formula follows from the one for $e^x$ .

$f(x)$	$f'(x)$	A check from the graph
$\sin x$	$\cos x$	slope 1 at 0, level at $\frac{\pi}{2}$
$\cos x$	$-\sin x$	level at 0, slope $-1$ at $\frac{\pi}{2}$
$e^x$	$e^x$	slope 1 at 0, slope equals height
$\ln x$	$\frac{1}{x}$ , for $x > 0$	slope 1 at $x = 1$ , flattening out

Worked example 1. Differentiate $f(x)$ $= 4e^x - 3\cos x + 5\ln x$ for $x > 0$ , and find $f'(\pi)$ .

Term by term, with the constant multiple and sum rules: $f'(x)$ $= 4e^x - 3(-\sin x)$ $+ \frac{5}{x}$ $= 4e^x$ $+ 3\sin x$ $+ \frac{5}{x}$ .

At $x = \pi$ : $f'(\pi)$ $= 4e^{\pi}$ $+ 3(0)$ $+ \frac{5}{\pi}$ $= 4e^{\pi}$ $+ \frac{5}{\pi}$ .

The step to watch is the double negative: $-3\cos x$ becomes $+3\sin x$ .

Worked example 2. Write the line tangent to $y = \ln x$ at $x = e$ .

Point: $\ln e = 1$ , so $(e, 1)$ . Slope: $\frac{1}{e}$ .

Line: $y = 1 + \frac{1}{e}(x - e)$ , which simplifies to $y = \frac{x}{e}$ . It passes through the origin.

Limits that are derivatives

Now that you know these derivatives, you can evaluate some limits instantly by reading them as derivatives at a point. This is the limit as a derivative idea, and it appears on the exam every year. The method:

Match the shape $\lim_{h \to 0} \frac{f(a + h) - f(a)}{h}$ or $\lim_{x \to a} \frac{f(x) - f(a)}{x - a}$ .
Name $f$ and $a$ . Check that the subtracted number really is $f(a)$ .
The limit is $f'(a)$ . Evaluate it with a derivative rule.

Worked example 3. Evaluate $\lim_{h \to 0} \frac{\sin\left( \frac{\pi}{6} + h \right) - \frac{1}{2}}{h}$ .

Shape: $f(x) = \sin x$ , $a = \frac{\pi}{6}$ , and $\sin \frac{\pi}{6} = \frac{1}{2}$ is indeed $f(a)$ .

So the limit is $f'\left( \frac{\pi}{6} \right)$ $= \cos \frac{\pi}{6}$ $= \frac{\sqrt{3}}{2}$ .

Worked example 4. Evaluate $\lim_{h \to 0} \frac{\ln(1 + h)}{h}$ .

The subtracted value is hidden: $\ln 1 = 0$ , so the numerator is $\ln(1 + h) - \ln 1$ . That is the shape with $f(x) = \ln x$ and $a = 1$ .

The limit is $f'(1) = \frac{1}{1} = 1$ .

Substituting $h = 0$ in these limits gives $\frac{0}{0}$ , which never settles anything. Recognizing the derivative does.

Common mistakes

Sign on cosine: $\frac{d}{dx}(\cos x) = -\sin x$ , not $\sin x$ .
Power rule on $e^x$ : the answer is $e^x$ .
Degrees: the trigonometric formulas need radians, on paper and on the calculator.
Answering $f(a)$ instead of $f'(a)$ when a limit is a derivative.

Sine and cosine

The exponential exe^x

The natural logarithm

Limits that are derivatives

Common mistakes

The exponential $e^x$