Differentiation: definition and properties

AVERAGE RATE, INSTANTANEOUS RATE, AND THE TWO DIFFERENCE QUOTIENTS

An average rate of change over an interval is a slope between two points on the graph — a difference quotient. The CED writes it two ways, and both appear on the exam:

They are the same fraction. In the first the interval runs from $a$ to $a+h$ and its width is $h$; in the second it runs from $a$ to $x$ and its width is $x-a$. Choose whichever makes the algebra cleaner and say which you chose.

The instantaneous rate of change at $a$ is what those averages approach as the interval shrinks onto $a$ — Unit 1’s opening question, now given a name and a symbol:

provided the limit exists. That proviso is topic 2.4, and it is not fine print.

Three sentences, one number. $f^{\prime }\left(a\right)$ is the slope of the tangent line at $x=a$; it is the instantaneous rate of change of $f$ at $a$; it is the limit of the average rates over intervals containing $a$. The exam asks for all three in words, and it asks for units: if $s\left(t\right)$ is feet and $t$ is seconds, then $s^{\prime }\left(2\right)$ is in feet per second, and the sentence “at $t=2$ the position is changing at $32$ ft/s” is the answer, not $32$.

THE DERIVATIVE AS A FUNCTION, AND THE FOUR WAYS TO WRITE IT

Do the limit at a general $x$ instead of a particular $a$ and the result is a new function, $f^{\prime }$:

Its value at any $x$ is the slope there. Here it is for $f\left(x\right)=x^2$, worked the only way Unit 2 allows until the rules are earned:

Substituting $h=0$ at the start gives $\frac{0}{0}$; canceling the $h$ first gives $2x+h$, which substitution can read. That is Unit 1’s factoring move, and every “use the definition” problem is exactly it.

Four notations, one object. For $y=f\left(x\right)$ the derivative is written $f^{\prime }\left(x\right)$, $y^{\prime }$, ${\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}$, or ${\displaystyle \frac{d}{\mathrm{dx}}}\left[f\right(x\left)\right]$. The last two are Leibniz’s, and they are worth understanding rather than tolerating: $\frac{\mathrm{dy}}{\mathrm{dx}}$ is $\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$ with the limit already taken — a slope written as a fraction whose parts have gone to zero together. $\frac{d}{\mathrm{dx}}$ on its own is an instruction (“take the derivative with respect to $x$”) waiting for something to act on. The exam uses all four without warning; the CED’s skill 4.C is literally “use appropriate mathematical symbols and notation.”

Four representations. A derivative can be given as a formula, as a slope on a graph, as a number estimated from a table, or as a sentence about a rate. The picture below is the one students misread most: the graph of $f^{\prime }$ is not a picture of $f$ — it is a picture of $f$’s slopes. Where $f$ rises, $f^{\prime }$ is positive; where $f$ has a horizontal tangent, $f^{\prime }$ is zero (it need not cross the axis); where $f$ falls, $f^{\prime }$ is negative.

THE TANGENT LINE — TWO NUMBERS AND A FORM

The tangent line at $x=a$ passes through the point $(a,f(a\left)\right)$ with slope $f^{\prime }\left(a\right)$. Point–slope form was built for exactly this:

Two numbers are all it needs, and they come from two different places: the point from $f$, the slope from $f^{\prime }$. Mixing them up — using $f\left(a\right)$ as the slope, or $f^{\prime }\left(a\right)$ as the height — is the whole error class on tangent-line questions. Leave the answer in point–slope form unless the question asks for another; rearranging is a place to drop a sign for no credit.

The normal line is perpendicular to the tangent at the same point: same point, slope $-{\displaystyle \frac{1}{f^{\prime }\left(a\right)}}$ (and vertical if $f^{\prime }\left(a\right)=0$).

Why the tangent matters beyond the question that asks for it. Zoom in on a differentiable point and the curve straightens — close enough, the function and its tangent line are the same line. That is what differentiability means geometrically (local linearity), and it is why Unit 4 will use the tangent line to approximate $f$ near $a$: $f\left(x\right)\approx f\left(a\right)+f^{\prime }\left(a\right)(x-a)$. The tangent line is the derivative made visible.

ESTIMATING A DERIVATIVE FROM A TABLE, A GRAPH, OR A CALCULATOR

When there is no formula there is no rule to apply, and the derivative is estimated from the average rate over the smallest interval the data allows.

From a table. For $f^{\prime }\left(c\right)$, take the two tabulated points that bracket $c$ most tightly and write the difference quotient with the function’s name in it:

That line is the point. The College Board says so directly: a correct number without this structure loses the credit. If $c$ is itself a tabulated point, use the points on either side of it (the symmetric difference $\frac{f(c+k)-f(c-k)}{2k}$) — it is a better estimate, and it is the one graders expect when both neighbors are available. Write $\approx$, never $=$: a table is evidence, not a formula.

From a graph. Draw the tangent by eye at the point, then find two places where your line crosses grid intersections and read rise over run between them. Below, the tangent at $(2,f(2\left)\right)$ passes through $\left(\mathrm{1,1}\right)$ and $\left(\mathrm{4,4}\right)$: rise $3$ over run $3$, so $f^{\prime }\left(2\right)\approx 1$. Say “approximately,” and expect a tolerance — the grader has one, and a line drawn by eye that reads $0.9$ or $1.1$ is a correct answer. What is not correct is the slope of a chord that starts at the point and ends somewhere on the curve; that is an average rate, and the question asked for the tangent’s.

With technology. A calculator’s numerical derivative (nDeriv, or $\frac{d}{\mathrm{dx}}$ at a value) is allowed on the calculator sections; the expression you evaluated is written on the page, and the result is rounded to three places after the decimal point unless told otherwise. Store intermediate values rather than retyping rounded ones.

DIFFERENTIABILITY AND CONTINUITY: WHEN A DERIVATIVE DOES NOT EXIST

Differentiable implies continuous. If $f^{\prime }\left(a\right)$ exists, then $f$ is continuous at $a$ — so a point not in the domain of $f$ is not in the domain of $f^{\prime }$, and a jump, a hole, or an asymptote at $a$ ends the question. The CED states this one direction only, and the direction matters: continuous does not imply differentiable. A function can pass Unit 1’s checklist at $a$ and still have no slope there, in three ways.

The test that scores. Differentiability at $a$ is a limit, so it is settled by one-sided limits of the difference quotient, exactly as a limit’s existence was settled in Unit 1:

must both exist and agree. For $\left|x\right|$ at $0$ the left quotient is $\frac{-h}{h}=-1$ and the right is $\frac{h}{h}=1$: a corner. For $\sqrt[3]{x}$ at $0$ the quotient is $\frac{h^{1∕3}}{h}=h^{-2∕3}$, which runs to $+\infty$ from both sides: a vertical tangent, the CED’s own example. A picture of a corner is not a justification; those two limits are.

Piecewise functions. To be differentiable at a seam a function must first be continuous there (Unit 1’s seam equation) and then have matching one-sided slopes — two equations, and “find $m$ and $b$ so that $f$ is differentiable at $1$” is asking for both. Solving only the slope equation, or only the continuity one, is half credit on a good day.

THE POWER RULE — AND THE REWRITING THAT HAS TO HAPPEN FIRST

For any real exponent $r$,

Bring the exponent down as a coefficient; subtract one from the exponent. It is the definition, computed once: for $x^2$ you saw $2x$ above; for $x^3$ the same expansion gives $\frac{3x^{2}h+3x{h}^2+h^3}{h}=3x^2+3\mathrm{xh}+h^2\to 3x^2$; and the pattern the binomial expansion produces is $r{x}^{r-1}$ every time, which is why one rule covers every power.

Rewrite before you differentiate. The rule reads $x^r$ and nothing else, so every root and every reciprocal is rewritten as a power first:

Then the rule, then (if the question wants it) rewrite back: $\frac{d}{\mathrm{dx}}\sqrt{x}=\frac{1}{2}{x}^{-1∕2}={\displaystyle \frac{1}{2\sqrt{x}}}$. Students who try to differentiate $\sqrt{x}$ “as a root” invent a rule; students who rewrite never need one.

The sign on a negative exponent. $\frac{d}{\mathrm{dx}}\left[x^{-3}\right]=-3x^{-4}$: the exponent goes more negative, from $-3$ to $-4$, because subtracting one from a negative number makes it smaller. Writing $-3x^{-2}$ is the standard slip, and it comes from “subtract one” being heard as “move toward zero.”

Constants. A constant has no slope: $\frac{d}{\mathrm{dx}}\left[7\right]=0$. A constant times $x$ has slope equal to the constant: $\frac{d}{\mathrm{dx}}\left[7x\right]=7$. And ${\pi }^2$, $e^3$, $\sqrt{5}$ are constants too, however much they look like functions — their derivatives are zero.

CONSTANT MULTIPLES, SUMS, AND DIFFERENCES — AND THE TWO RULES THAT DO NOT EXIST

Derivatives pass through addition, subtraction, and constant multiples:

These are Unit 1’s limit laws wearing new clothes: a limit walks through a sum and through a constant factor, and the derivative is a limit. So a polynomial is differentiated term by term, each term by the power rule, each coefficient carried along:

The two rules that do not exist. The derivative of a product is not the product of the derivatives, and the derivative of a composition is not the derivative of the outside evaluated at the inside. Test the first on $x\cdot x$: the true derivative of $x^2$ is $2x$, and the product of the derivatives is $1\cdot 1=1$. Wrong on the simplest case there is. The product rule (two cards down) is what the limit actually produces; the chain rule for compositions is Unit 3’s, and until then ${(2x+1)}^5$ is handled by expanding it or by waiting — never by “$5{(2x+1)}^4$” alone, which is the chain rule with its second factor missing and is marked as such.

THE FOUR MEMORIZED DERIVATIVES: $\sin x$, $\cos x$, $e^x$, $\ln x$

Four facts, and the sign on the second is the one that leaves: cosine’s derivative is negative sine. Say it aloud when you write it.

Where the sine rule comes from — because a rule you can rebuild is a rule you cannot lose. Apply the definition and the angle-addition identity $\sin (x+h)=\sin x\cos h+\cos x\sin h$:

As $h\to 0$ the first fraction goes to $0$ and the second to $1$ — Unit 1’s two memorized trig limits, which were never really about limits. They were the derivative of sine at zero, waiting. So the whole thing goes to $\sin x\cdot 0+\cos x\cdot 1=\cos x$.

Where the exponential rule comes from. $\frac{e^{x+h}-e^x}{h}=e^x\cdot \frac{e^h-1}{h}$, and the second factor goes to $1$ — check it: $h=0.01$ gives $\frac{1.01005-1}{0.01}=1.005$. That limit being exactly $1$ is what makes $e$ the base calculus prefers: the function whose rate of change is itself. The logarithm’s rule waits for Unit 3, where it falls out of $e^x$ by the inverse relationship; until then it is a fact to carry, with its domain — $\ln x$ lives on $x>0$, so $\frac{1}{x}$ as its derivative means the positive branch only.

A LIMIT THAT IS A DERIVATIVE IN DISGUISE

The CED’s topic 2.7 carries a second learning objective the title hides (LIM-3.A): recognize the definition of the derivative when a limit is written in its shape, and evaluate the limit by a rule instead of by algebra. The tell is the form:

Find the $f$ and the $a$. In $\underset{h\to 0}{\lim }\frac{\sin \left(\frac{\pi }{6}+h\right)-\frac{1}{2}}{h}$ the function is $\sin$, the point is $\frac{\pi }{6}$ (and $\frac{1}{2}$ is $\sin \frac{\pi }{6}$, which is the confirmation), so the limit is $\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}$. No algebra, no identity — a rule. In $\underset{x\to 2}{\lim }\frac{x^5-32}{x-2}$ the function is $x^5$ at $a=2$, and the answer is $5\cdot 2^4=80$; try that one by factoring and you will see why recognizing is worth learning.

The trap is a limit that almost has the shape: $\frac{f(a+h)-f(a-h)}{h}$ is $2f^{\prime }\left(a\right)$, not $f^{\prime }\left(a\right)$; $\frac{f(a+h)-f\left(a\right)}{2h}$ is $\frac{1}{2}{f}^{\prime }\left(a\right)$. Match the pieces exactly before you name the derivative.

THE PRODUCT RULE — WHY IT HAS TWO TERMS

Each factor gets its turn to change while the other holds still, and the two contributions add. The picture makes it obvious. Let $u$ and $v$ be the sides of a rectangle, so $\mathrm{uv}$ is its area; grow $x$ a little and each side grows a little:

A product changes for two reasons — $u$ moves, and $v$ moves — so its derivative has two terms, one for each. The corner is the piece that is neither, and it is worth being exact about how it leaves. Divide the area change by $h$ and the corner becomes $\mathrm{\Delta }u\mathrm{\Delta }v∕h=(\mathrm{\Delta }u∕h)\mathrm{\Delta }v$; for differentiable factors the first part goes to $u^{\prime }$ while the second goes to $0$, so the product goes to $0$. Note what that argument does not say. It does not say $u^{\prime }{v}^{\prime }$ vanishes — $u^{\prime }{v}^{\prime }$ is generally a perfectly good nonzero number, and on $u=v=x$ it is $1$. It says only that $u^{\prime }{v}^{\prime }$ is not the rate of change of $\mathrm{uv}$, which is the whole reason the wrong rule is wrong. The College Board names this error in the unit’s own overview; it is the most common wrong derivative on the exam that is not an arithmetic slip.

The structure before the values. On the exam, a product rule at a point is written as a structure first — $f^{\prime }\left(3\right)=u^{\prime }\left(3\right)v\left(3\right)+u\left(3\right)v^{\prime }\left(3\right)$ — and only then are the four numbers imported from the table or the graph. The structure is the point; the arithmetic is the answer. Both are needed, and they are scored separately.

When not to use it. $x^2(x+1)$ is a product, but it is also $x^3+x^2$, and the power rule on the expanded form is faster and safer. Expand when the factors are polynomials; use the rule when one factor is $\sin x$, $e^x$, $\ln x$, or anything that will not multiply out.

THE QUOTIENT RULE — AND THE SIGN THAT HAS A SIDE

The numerator is the product rule’s two terms with a minus between them, and unlike the product rule, the order matters: the derivative of the top comes first. Swap them and the sign of the whole derivative flips — an error that survives every later step and produces a tangent line sloping the wrong way. The denominator is squared, and it is the original denominator, not its derivative.

The self-check that never fails. Run the rule on $\frac{1}{x}$: $u=1$, $v=x$, so the derivative is $\frac{0\cdot x-1\cdot 1}{x^2}=-\frac{1}{x^2}$. The power rule on $x^{-1}$ gives $-x^{-2}$, the same thing. If the quotient rule in your head gives $+\frac{1}{x^2}$ here, the order is backwards and you have found out for free.

When to avoid it. A quotient whose bottom is a single power of $x$ is a sum of powers in disguise: $\frac{x^2+1}{x^3}=x^{-1}+x^{-3}$, and the power rule term by term is shorter than the quotient rule and has no sign to get wrong. Problem 10 does it both ways so you can watch them agree.

TANGENT, COTANGENT, SECANT, COSECANT — FOUR QUOTIENTS, ONE METHOD

Nothing new is memorized here unless you want it to be. Each of the four is a quotient of sine and cosine, and the quotient rule produces its derivative on demand:

The other three go the same way. The results, for reference:

The pattern is worth one sentence: every co-function’s derivative carries a minus sign ($\cos$, $\cot$, $\csc$), and every derivative comes in the same family it started in. A student who remembers that sentence can rebuild the table from $\tan x\to {\sec }^{2}x$ in a minute; a student who remembers only the table has four signs to lose.