# What is a derivative? Speed at a single instant > A falling ball has a speed at every instant, but an instant is too short to measure a speed over. The derivative is the precise idea that resolves this. Series: Calculus (https://www.monocurl.com/monocurl-essays/calculus/) Canonical: https://www.monocurl.com/monocurl-essays/calculus/what-is-a-derivative/ Keywords: derivative, definition of the derivative, instantaneous rate of change, instantaneous velocity, difference quotient, tangent line, limit, differentiable [Figure 1, animation: A ball dropped from rest, in real time. Its distance fallen after t seconds is 5t^2 meters, and the curve on the left records that distance as it grows.] Drop a ball and let it fall for two seconds. When you let go it is hardly moving, and by the end it is moving fast enough to hurt. So at each moment in between it had some definite speed, and in particular it had one exactly one second after release. What was it? The question sounds routine, but try to answer it with the only definition of speed you have. Speed is distance divided by time, which means that measuring it takes two moments: where the ball was then, and where it is now. "Exactly one second after release" is a single moment. During a single moment the ball covers no distance and no time passes, so the recipe asks us to divide zero meters by zero seconds, and that division has no answer. Yet nobody doubts that the ball has a speed at that moment. A radar gun would report one, and if the ball landed in your hand at that instant, how much it stung would depend on it. So we are sure that a quantity exists while our definition of it refuses to produce a value. The way out is the first real idea of calculus. Newton and Leibniz found it in the seventeenth century, and it took well over a century more to state it in a form that survives every objection. ## Average speed is something we can compute We need a description of the fall to work with. Ignoring air resistance, a ball dropped from rest near the Earth's surface has fallen approximately $$ s(t) = 5t^2 \text{ meters} $$ after $t$ seconds. (The true coefficient is closer to 4.9; rounding keeps the arithmetic simple.) One second in, the ball has fallen $s(1) = 5$ meters. Two seconds in, it has fallen $s(2) = 20$ meters. The animation above plots this function: time runs along the bottom, distance fallen goes up the side. Speed at one moment is out of reach for now, but speed over an interval of time is ordinary arithmetic. Between $t = 1$ and $t = 2$ the ball goes from 5 meters to 20 meters, so it covers 15 meters in 1 second: $$ \text{average speed from } t = 1 \text{ to } t = 2 \;=\; \frac{s(2) - s(1)}{2 - 1} = \frac{20 - 5}{1} = 15 \text{ m/s}. $$ This is a true statement about one second of the fall and a poor answer to our question, because the ball was speeding up during that whole second. At $t = 1$ it must have been moving slower than 15 m/s, and at $t = 2$ faster. The average is contaminated by everything that happened after the moment we care about. If the trouble is that the speed changes during the interval, the repair is a shorter interval, in which it has less time to change. [Figure 2, steps: Average speed over a shrinking interval of time that starts at t = 1. The orange line joins the ball's position at the two ends of the interval, and its slope is the average speed.] - step: From 1 s to 2 s the ball falls 15 m. Average speed: 15 m/s. - step: Halve the stretch. From 1 s to 1.5 s: 6.25 m in half a second, or 12.5 m/s. - step: A tenth of a second: 10.5 m/s. - step: A hundredth of a second: 10.05 m/s. The two points are now too close to tell apart. Over intervals of length 1, 0.5, 0.1 and 0.01 seconds, all starting at $t = 1$, the average speeds in the readout are 15, 12.5, 10.5 and 10.05 meters per second. **Pause and try.** Compute the average speed over the interval from $t = 1$ to $t = 1.001$. Before you calculate, write down a guess based on the four values above. (You will need $1.001^2 = 1.002001$.) **Answer.** At the end of the interval the ball has fallen $s(1.001) = 5 \times 1.002001 = 5.010005$ meters, so it covers $0.010005$ meters in $0.001$ seconds. The average speed is $0.010005 / 0.001 = 10.005$ m/s. Each time the interval shrinks by a factor of ten, the amount by which the average exceeds 10 shrinks by a factor of ten as well: $0.5$, then $0.05$, then $0.005$. ## One formula for every interval The averages are plainly heading for 10. Rather than compute more of them, we can find out why by letting the length of the interval be a variable. Call that length $h$, and take $h \neq 0$. The interval runs from time $1$ to time $1 + h$, and the average speed over it is the distance covered divided by the time taken: $$ q(h) = \frac{s(1 + h) - s(1)}{h} = \frac{5(1+h)^2 - 5}{h}. $$ Expand the square, $(1 + h)^2 = 1 + 2h + h^2$, so the numerator is $5 + 10h + 5h^2 - 5 = 10h + 5h^2$. Both terms that remain contain a factor of $h$, and because $h \neq 0$ we are allowed to divide it out: $$ q(h) = \frac{10h + 5h^2}{h} = 10 + 5h \qquad (h \neq 0). $$ This one line reproduces the whole table: $q(1) = 15$, $q(0.5) = 12.5$, $q(0.1) = 10.5$, $q(0.01) = 10.05$, and $q(0.001) = 10.005$ as you found above. Nothing in the calculation required $h$ to be positive. A negative $h$ describes an interval that *ends* at $t = 1$ instead of starting there. For instance $h = -0.1$ is the interval from $t = 0.9$ to $t = 1$, over which the formula gives an average speed of $q(-0.1) = 9.5$ m/s. That is below 10, as it should be, since the ball was slower before $t = 1$ than after. [Figure 3, image: The average speed q(h) = 10 + 5h plotted against the interval length h, for intervals after t = 1 (h > 0) and before it (h < 0). The function q is not defined at h = 0, which is the open circle, and its values close in on 10 from both sides.] The graph of $q$ is a straight line with one point missing, and the missing point is the one we care about. The expression $10 + 5h$ makes sense at $h = 0$, where it equals 10. But $q(h)$ was defined as a quotient with $h$ in the denominator, and dividing $h$ out was only legitimate for $h \neq 0$, so $q(0)$ does not exist and no simplifying afterwards can bring it into existence. What the simplified formula describes is all the *other* values: every average speed over an interval with one end at $t = 1$ equals 10 plus a leftover of $5h$, and that leftover can be made as small as we please by taking the interval short enough. **Pause and try.** Suppose you want the average speed to be within $0.001$ m/s of 10. How short must the interval be? Does it matter whether the interval comes before or after $t = 1$? **Answer.** We need $|q(h) - 10| < 0.001$. Since $q(h) - 10 = 5h$, this says $5|h| < 0.001$, that is, $|h| < 0.0002$. Any interval shorter than two ten-thousandths of a second will do, and it makes no difference on which side of $t = 1$ it lies, because only the size of $h$ entered the calculation. ## What "closes in on" means That exercise contains the whole idea. We claimed the averages "close in on" 10, and a skeptic is entitled to ask what that phrase commits us to. It cannot mean that $q(h)$ ever equals 10, because it never does. It cannot merely mean that $q(h)$ gets closer to 10 as a positive $h$ shrinks, because then $q(h)$ also gets closer to 9, and to 0, without those numbers having any claim on it. What sets 10 apart is that we can meet *any* demand for closeness. Name a tolerance, however small, and there is a length such that every interval shorter than it has an average speed within your tolerance of 10. The number 9 fails this test: every interval shorter than a tenth of a second has an average speed above 9.5, so a tolerance of 0.5 around 9 cannot be met. Written with the traditional letters, $\varepsilon$ for the tolerance and $\delta$ for the length, this becomes the definition of a limit, which has [a lesson of its own](@calculus/limits). **Definition.** Let $q$ be a function defined for all $h \neq 0$ in some open interval around 0. We say that **$q(h)$ has limit $L$ as $h$ approaches 0**, and write $$ \lim_{h \to 0} q(h) = L, $$ if for every number $\varepsilon > 0$ there is a number $\delta > 0$ such that $$ 0 < |h| < \delta \quad \text{implies} \quad |q(h) - L| < \varepsilon. $$ Read the clauses against what we did. "For every $\varepsilon > 0$" is the skeptic choosing a tolerance. "There is a $\delta > 0$" is our reply, a length that works for that tolerance. The condition $0 < |h|$ excludes $h = 0$, so the definition never asks about $q(0)$ and does not care whether it exists. And $|h| < \delta$ allows $h$ of either sign, so intervals on both sides of the instant are required to agree. For our function the verification is a single line. Given $\varepsilon > 0$, choose $\delta = \varepsilon / 5$. If $0 < |h| < \delta$, then $|q(h) - 10| = |5h| = 5|h| < 5\delta = \varepsilon$. Hence $$ \lim_{h \to 0} q(h) = 10. $$ One more fact justifies the phrase "the limit": a function cannot have two different limits at the same point. If it had limits $L \neq L'$, take $\varepsilon$ to be half the distance between them. Sufficiently small $h$ would put $q(h)$ within $\varepsilon$ of both, which is impossible because the intervals of radius $\varepsilon$ around $L$ and $L'$ do not overlap. ## The definition of the derivative We can now say what the speed of the ball at the instant $t = 1$ *is*. It is 10 meters per second, and the meaning of that statement is: 10 is the limit of the average speeds over intervals with an end at $t = 1$, as the length of the interval approaches zero. We never measure at the instant itself; we measure across short intervals that end there and ask what number those measurements pin down. You might object that this is a dodge, since we wanted the speed *at* the instant and have settled for a statement about nearby intervals. But a radar gun, a speedometer and a stinging hand all respond to what the ball does over some very short interval of time; none of them has access to a single instant. "The number that all sufficiently short intervals agree on, to within any tolerance" is the only thing a speed at an instant could mean, and the definition says so openly. Nothing in the construction was specific to falling or to speed, so it can be stated for any function. **Definition.** Let $f$ be a function defined on an open interval containing the number $a$. For $h \neq 0$ small enough that $a + h$ is in that interval, the **difference quotient** of $f$ at $a$ is $$ \frac{f(a+h) - f(a)}{h}. $$ If the difference quotient has a limit as $h$ approaches 0, we say $f$ is **differentiable at $a$**, and we call the limit the **derivative of $f$ at $a$**: $$ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. $$ The difference quotient is an average rate of change: the change in the output, $f(a+h) - f(a)$, divided by the change in the input, $h$. The derivative is the instantaneous rate of change. For the ball, $f$ is $s$, the input is time, the output is distance, and the derivative is the speed. The "if" in the definition leaves open the possibility that the limit does not exist, in which case there is no derivative at that point; we will meet such a function below. Nor is there any reason to stop at $t = 1$. **Theorem.** The function $s(t) = 5t^2$ is differentiable at every real number $a$, and $s'(a) = 10a$. **Proof.** Fix a real number $a$ and let $h \neq 0$. The difference quotient of $s$ at $a$ is $$ \frac{5(a+h)^2 - 5a^2}{h} = \frac{5a^2 + 10ah + 5h^2 - 5a^2}{h} = \frac{10ah + 5h^2}{h} = 10a + 5h, $$ where the last step divides by $h$, which is not zero. We claim this has limit $10a$ as $h \to 0$. Let $\varepsilon > 0$ and choose $\delta = \varepsilon / 5$. If $0 < |h| < \delta$, then $$ |(10a + 5h) - 10a| = 5|h| < 5\delta = \varepsilon. $$ So the limit exists and equals $10a$, which is to say $s'(a) = 10a$. At $a = 1$ the theorem gives 10, as before. At $a = 2$ it gives 20, and at $a = 0$, the moment of release, it gives 0. The ball gains 10 meters per second of speed for each second it falls. Galileo reached that law through experiments with balls rolling down ramps; here it has come out of the distance formula by algebra alone. [Figure 4, animation: The tangent line rides along the distance curve (left). Plot its slope at each moment and you get the ball's speed (right): the straight line 10t.] We began with one function, the distance $s(t) = 5t^2$, and by asking the same question at every instant we produced a second function, the speed $s'(t) = 10t$. The second function is called **the derivative** of the first, and the process of finding it is called **differentiation**. ## The derivative as a slope There is a second way to see the number 10, and it is the picture most people carry in their heads. On the graph of $s$, an average speed is a [slope](@algebra/slope-and-lines). Pick two points on the curve, at times $1$ and $1 + h$. The distance covered between them is the vertical separation of the points and the time elapsed is the horizontal separation, so the average speed is rise over run: the slope of the straight line through the two points. Such a line is called a **secant line**, and the difference quotient is its slope. Shrinking the interval slides the second point along the curve toward the first. The secant line pivots as it goes, and its slope has a limit. [Figure 5, animation: As the second point slides toward the first, the secant line through both settles onto the tangent line at t = 1, and its slope settles at 10.] The line through the point $(1, 5)$ whose slope is that limit is called the tangent line. The definition is general. **Definition.** If $f$ is differentiable at $a$, the **tangent line** to the graph of $f$ at the point $(a, f(a))$ is the line through that point with slope $f'(a)$. Its equation is $$ y = f(a) + f'(a)(x - a). $$ For the ball at $a = 1$ the tangent line is $y = 5 + 10(t - 1)$. Notice that the definition does not say "the line that touches the curve at one point". That description, which works for circles, is neither precise nor true in general: a tangent line can cross the curve elsewhere, and at some points it even crosses the curve at the point of tangency. So the derivative has two readings, and they are the same number: the rate at which the output is changing at one input, and the slope of the graph at one point. The second reading explains why the first is reasonable. Zoom in on the graph at $(1, 5)$, with a grid so that slopes can be read off. [Figure 6, animation: Zooming in on the graph of s at the point (1, 5). The grid rescales with the zoom, and every cell is ten times as many meters tall as it is seconds wide. Up close the curve climbs one cell for each cell it crosses: a straight line of slope 10.] Under magnification the curve becomes indistinguishable from a straight line, and the grid measures that line: one cell up for each cell across, with cells ten times as many meters tall as seconds wide, so the slope is 10 meters per second. A curve that looks straight over a tiny interval is behaving, over that interval, like an object moving at a steady speed. The tangent line is that steady motion continued in both directions. We can make "indistinguishable" precise with what we already know. The gap between the curve and its tangent line at time $1 + h$ is $$ s(1+h) - \bigl(5 + 10h\bigr) = 5 + 10h + 5h^2 - 5 - 10h = 5h^2. $$ At $h = 0.1$ the gap is 0.05 meters. At $h = 0.01$ it is 0.0005 meters. When the horizontal distance from the point shrinks by a factor of 10, the gap shrinks by a factor of 100. That is why zooming flattens the curve onto the line: the screen is magnified in proportion to $1/h$, while the gap shrinks like $h^2$. **Pause and try.** The tangent line predicts the ball's position a short time after $t = 1$. Use it to estimate $s(1.2)$, then compute the true value. How large is the error, and does it match the formula $5h^2$? **Answer.** The tangent line gives $5 + 10(0.2) = 7$ meters. The true value is $s(1.2) = 5 \times 1.44 = 7.2$ meters. The error is $0.2$ meters, and $5h^2 = 5 \times 0.04 = 0.2$. The tangent line underestimates because the ball keeps accelerating after $t = 1$. ## Notation Two notations for the derivative are in common use. The prime notation, usually credited to Lagrange, writes the derivative of $f$ as $f'$, so that $f'(a)$ is its value at $a$. It is compact and makes clear that the derivative is itself a function. The notation of Leibniz starts from the difference quotient. Write $\Delta t$ for a change in $t$ (the letter $\Delta$ is read "change in") and $\Delta s$ for the corresponding change in $s$. An average rate is then $\Delta s / \Delta t$, and for its limit Leibniz wrote $$ \frac{ds}{dt} = \lim_{\Delta t \to 0} \frac{\Delta s}{\Delta t}. $$ So $\frac{ds}{dt} = 10t$ says the same thing as $s'(t) = 10t$. The Leibniz form keeps the units visible, since meters over seconds gives meters per second, and it will make several later rules easy to remember. But $\frac{ds}{dt}$ is one symbol for a limit, and $ds$ and $dt$ are not two numbers being divided: in the definition nothing is ever divided by zero, and nothing "infinitely small" appears. ## Worked examples The definition is also a method: write the difference quotient, simplify it until the $h$ in the denominator cancels, and then take the limit. **Worked example.** **The derivative of $f(x) = x^2$ at $x = 3$, and the tangent line there.** The difference quotient at $a = 3$ is $$ \frac{(3+h)^2 - 9}{h} = \frac{9 + 6h + h^2 - 9}{h} = \frac{6h + h^2}{h} = 6 + h \qquad (h \neq 0). $$ As $h \to 0$ this has limit 6: given $\varepsilon > 0$, take $\delta = \varepsilon$, and then $0 < |h| < \delta$ gives $|(6 + h) - 6| = |h| < \varepsilon$. So $f'(3) = 6$. The tangent line passes through $(3, 9)$ with slope 6, so its equation is $y = 9 + 6(x - 3)$, or $y = 6x - 9$. *Check.* At $x = 3.1$ the line gives $9.6$ and the function gives $3.1^2 = 9.61$. They differ by $0.01 = h^2$, the same kind of second-order gap we saw for the ball. **Worked example.** **The derivative of $f(x) = \dfrac{1}{x}$ at a point $a \neq 0$.** Here the difference quotient is a fraction built out of fractions, and the way to simplify it is to combine the two terms of the numerator over a common denominator: $$ \frac{1}{a+h} - \frac{1}{a} = \frac{a - (a + h)}{a(a+h)} = \frac{-h}{a(a+h)}. $$ Dividing by $h$, $$ \frac{f(a+h) - f(a)}{h} = \frac{-1}{a(a+h)} \qquad (h \neq 0,\ a + h \neq 0). $$ As $h \to 0$ the factor $a + h$ approaches $a$, so the quotient approaches $-1/a^2$. (We are using the fact that the limit of a quotient is the quotient of the limits when the denominator's limit is not zero. That is one of the limit laws, which are proved from the $\varepsilon$–$\delta$ definition in the [lesson on limits](@calculus/limits).) Therefore $$ f'(a) = -\frac{1}{a^2}. $$ The derivative is negative at every point, which matches the graph of $1/x$: it falls as you move right, on both sides of zero. A derivative carries a sign, and a negative derivative means the output is decreasing. **Worked example.** **The derivative of $f(x) = x^3$ at any $a$, with the limit verified from the definition.** Expanding $(a+h)^3 = a^3 + 3a^2h + 3ah^2 + h^3$, the difference quotient is $$ \frac{3a^2h + 3ah^2 + h^3}{h} = 3a^2 + 3ah + h^2 \qquad (h \neq 0). $$ We expect the limit $3a^2$. The leftover is $3ah + h^2 = h(3a + h)$, and this time it is not simply a multiple of $h$, so choosing $\delta$ takes a little care. First insist that $|h| < 1$. Then $|3a + h| \le 3|a| + |h| < 3|a| + 1$, so $$ |h(3a + h)| < |h|\,(3|a| + 1). $$ Given $\varepsilon > 0$, choose $\delta$ to be the smaller of $1$ and $\varepsilon / (3|a| + 1)$. If $0 < |h| < \delta$, both requirements hold, and the leftover is less than $\varepsilon$. Hence $f'(a) = 3a^2$. Put the three results side by side: the derivative of $x^2$ is $2x$, of $x^3$ is $3x^2$, and of $x^{-1}$ is $-x^{-2}$. There is a pattern, and [the next lesson](@calculus/power-rule) explains it. ## When there is no derivative The definition said "if the limit exists". Here is the standard example in which it does not. Let $f(x) = |x|$, the absolute value of $x$, and look at $a = 0$. The difference quotient is $$ \frac{|0 + h| - |0|}{h} = \frac{|h|}{h}, $$ which equals $1$ whenever $h > 0$ and $-1$ whenever $h < 0$. [Figure 7, steps: The function f(x) = |x| at x = 0. Secant lines to the right of the corner all have slope 1 and secant lines to the left all have slope -1, so the difference quotient has no single limit and f has no derivative at 0.] - step: Take a second point to the right of 0. The secant line through it and the corner has slope 1. - step: Slide that point toward the corner. The slope does not change: it is 1 for every positive h. - step: Now approach from the left. Every secant line there has slope -1. No number $L$ can be the limit. If one were, then taking $\varepsilon = 1$ would give a $\delta$ such that every $h$ with $0 < |h| < \delta$ has $\left|\frac{|h|}{h} - L\right| < 1$. A small positive $h$ would force $|1 - L| < 1$, so $L > 0$; a small negative $h$ would force $|-1 - L| < 1$, so $L < 0$. Both cannot hold. So $|x|$ is not differentiable at 0, and its graph has no tangent line there. Geometrically the graph has a corner, and zooming in on a corner never straightens it out. A break in the graph is another way to fail, and the following theorem shows that it always destroys the derivative. **Theorem.** If $f$ is differentiable at $a$, then $f$ is continuous at $a$; that is, $\lim_{h \to 0} f(a + h) = f(a)$. **Proof.** For $h \neq 0$ we can write $$ f(a+h) - f(a) = \frac{f(a+h) - f(a)}{h} \cdot h. $$ As $h \to 0$, the first factor has limit $f'(a)$ by hypothesis, and the second factor has limit 0. The limit of a product is the product of the limits (another of the limit laws), so $f(a+h) - f(a)$ has limit $f'(a) \cdot 0 = 0$, which is the claim. The converse is false, and $|x|$ is the witness: it is continuous at 0 and not differentiable there. Differentiability asks that the graph be unbroken and, in addition, that it have a single well-defined direction at the point. **Common mistake.** **Setting $h = 0$ too early.** It is tempting to "evaluate" the difference quotient by putting $h = 0$ straight into $\frac{5(1+h)^2 - 5}{h}$. That produces $\frac{0}{0}$, which is undefined, and no conclusion follows from it; in particular it does not show that the derivative fails to exist. The order of operations is: simplify while $h \neq 0$ until the $h$ in the denominator has cancelled, and only then ask for the limit. **Common mistake.** **Using $f(a)/a$ for the rate.** At $t = 1$ the ball has fallen 5 meters in 1 second, and many students answer that its speed is therefore 5 m/s. That quotient is the average speed over the interval from $t = 0$ to $t = 1$, during most of which the ball was moving slowly. The speed at $t = 1$ is 10 m/s, twice as large. Total over total gives an average over the whole history; the derivative describes what is happening now. **Check yourself.** For a different falling object, the distance fallen after $t$ seconds is $s(t) = 3t^2$ meters. What is its speed at $t = 2$? Answer: 12 m/s The difference quotient at $t = 2$ is $\frac{3(2+h)^2 - 12}{h} = 12 + 3h$, which has limit 12 as $h \to 0$. **Recap.** - Average rate of change over an interval is the difference quotient $\frac{f(a+h) - f(a)}{h}$, defined for $h \neq 0$. On the graph it is the slope of a secant line. - $\lim_{h \to 0} q(h) = L$ means: for every $\varepsilon > 0$ there is a $\delta > 0$ such that $0 < |h| < \delta$ implies $|q(h) - L| < \varepsilon$. The value at $h = 0$ plays no part. - The derivative $f'(a)$ is the limit of the difference quotient as $h \to 0$, when that limit exists. It is the instantaneous rate of change, and the slope of the tangent line $y = f(a) + f'(a)(x - a)$. - To compute one from the definition: form the quotient, simplify until $h$ cancels from the denominator, then take the limit. - For $s(t) = 5t^2$, $s'(t) = 10t$. Also $(x^2)' = 2x$, $(x^3)' = 3x^2$ and $(1/x)' = -1/x^2$. - Differentiable implies continuous, but not conversely: $|x|$ is continuous at 0 and has no derivative there. ## Problems Work these with pencil and paper before opening the solutions. The first three repeat the moves of the lesson, the next four ask you to combine them, and the last two go a little beyond it. **Problem 1.** For the ball with $s(t) = 5t^2$, find the average speed over the interval from $t = 2$ to $t = 2.5$, and the speed at the instant $t = 2$. Which is larger, and why must it be? **Solution.** $s(2) = 20$ and $s(2.5) = 5 \times 6.25 = 31.25$, so the average speed is $\frac{31.25 - 20}{0.5} = 22.5$ m/s. The speed at $t = 2$ is $s'(2) = 10 \times 2 = 20$ m/s. The average is larger because the ball speeds up throughout the interval, so at every later moment in it the ball is moving faster than it was at the start. **Problem 2.** Let $f(x) = 7x - 4$. Compute $f'(a)$ from the definition. Explain the answer in terms of the graph of $f$. **Solution.** The difference quotient is $\frac{7(a+h) - 4 - (7a - 4)}{h} = \frac{7h}{h} = 7$ for every $h \neq 0$. A function that is constantly 7 has limit 7 (any $\delta$ works for any $\varepsilon$), so $f'(a) = 7$ for all $a$. The graph is a straight line of slope 7. Every secant line of a straight line is the line itself, and so is the tangent line at each point. **Problem 3.** Find the equation of the tangent line to $y = x^2$ at the point $(-1, 1)$. **Solution.** The difference quotient of $x^2$ at $a$ is $\frac{(a+h)^2 - a^2}{h} = 2a + h$, with limit $2a$. At $a = -1$ the slope is $-2$. The line through $(-1, 1)$ with slope $-2$ is $y = 1 - 2(x + 1)$, that is, $y = -2x - 1$. Check: at $x = -1$ it gives $2 - 1 = 1$, as required. **Problem 4.** Let $f(x) = x^2 + 3x$. Find $f'(a)$ from the definition, including an $\varepsilon$–$\delta$ verification of the limit. **Solution.** $f(a+h) - f(a) = a^2 + 2ah + h^2 + 3a + 3h - a^2 - 3a = 2ah + h^2 + 3h$. Dividing by $h \neq 0$ gives $2a + 3 + h$. Given $\varepsilon > 0$, let $\delta = \varepsilon$; if $0 < |h| < \delta$ then $|(2a + 3 + h) - (2a + 3)| = |h| < \varepsilon$. So $f'(a) = 2a + 3$. Observe that this is the derivative of $x^2$ plus the derivative of $3x$. **Problem 5.** On the Moon a dropped object falls approximately $s(t) = 0.8t^2$ meters in $t$ seconds. Find its speed at time $t$. When is it falling at 8 m/s, and how far has it fallen by then? **Solution.** The difference quotient at $a$ is $\frac{0.8(a+h)^2 - 0.8a^2}{h} = 1.6a + 0.8h$, with limit $1.6a$. So the speed at time $t$ is $1.6t$ m/s. It equals 8 when $t = 5$ seconds, at which point the object has fallen $s(5) = 0.8 \times 25 = 20$ meters. **Problem 6.** Using only the $\varepsilon$–$\delta$ definition, prove that $\lim_{h \to 0} (3 - 2h) = 3$. Then explain why the same function does **not** have limit $3.1$. **Solution.** Given $\varepsilon > 0$, let $\delta = \varepsilon / 2$. If $0 < |h| < \delta$ then $|(3 - 2h) - 3| = 2|h| < 2\delta = \varepsilon$. For the second part, take $\varepsilon = 0.05$. Whatever $\delta > 0$ is proposed, choose $h$ with $0 < |h| < \delta$ and also $|h| < 0.01$. Then $3 - 2h$ lies between $2.98$ and $3.02$, so its distance from $3.1$ is at least $0.08$, which is not less than $0.05$. No $\delta$ works for this $\varepsilon$, so $3.1$ is not the limit. **Problem 7.** Let $f(x) = \sqrt{x}$ and let $a > 0$. Show that the difference quotient equals $\dfrac{1}{\sqrt{a+h} + \sqrt{a}}$, and deduce $f'(a)$. What happens to the quotient at $a = 0$ as $h \to 0$ through positive values? **Solution.** Multiply numerator and denominator by $\sqrt{a+h} + \sqrt{a}$: $$ \frac{\sqrt{a+h} - \sqrt{a}}{h} = \frac{(a + h) - a}{h\,(\sqrt{a+h} + \sqrt{a})} = \frac{1}{\sqrt{a+h} + \sqrt{a}}. $$ As $h \to 0$, $\sqrt{a+h}$ approaches $\sqrt{a}$ (the square root function is continuous), so the quotient approaches $\frac{1}{2\sqrt{a}}$. Hence $f'(a) = \frac{1}{2\sqrt{a}}$ for $a > 0$. At $a = 0$ the quotient is $\frac{\sqrt{h}}{h} = \frac{1}{\sqrt{h}}$, which grows without bound as $h \to 0^+$: for $h = 0.0001$ it is already 100. So there is no limit, and $\sqrt{x}$ is not differentiable at 0. The graph is vertical there. **Problem 8.** The function $f(x) = x\,|x|$ involves an absolute value. Is it differentiable at 0? Decide from the definition. **Solution.** The difference quotient at 0 is $\frac{h|h| - 0}{h} = |h|$ for $h \neq 0$. Given $\varepsilon > 0$, take $\delta = \varepsilon$: if $0 < |h| < \delta$ then $\bigl||h| - 0\bigr| = |h| < \varepsilon$. So the limit exists and $f'(0) = 0$. Its graph is $x^2$ for $x \ge 0$ and $-x^2$ for $x < 0$, two parabolic arcs that meet at the origin with the same slope, 0, so there is no corner. Whether a derivative exists is settled by the limit, not by the look of the formula. **Problem 9.** Some calculators estimate derivatives with the *symmetric* quotient $\dfrac{f(a+h) - f(a-h)}{2h}$. (a) Compute it for $s(t) = 5t^2$ and compare with $s'(a)$. (b) Compute it for $f(x) = |x|$ at $a = 0$. (c) What does (b) tell you about using the symmetric quotient as a definition of the derivative? **Solution.** (a) $5(a+h)^2 - 5(a-h)^2 = 5\,(4ah) = 20ah$, so the symmetric quotient is $\frac{20ah}{2h} = 10a$ exactly, for every $h \neq 0$. It equals $s'(a)$ with no leftover at all, which is why it is popular for numerical work. (b) $\frac{|h| - |-h|}{2h} = \frac{0}{2h} = 0$ for every $h \neq 0$, so its limit is 0. (c) The symmetric quotient has a limit at a point where $|x|$ has no derivative. It averages the slopes on the two sides, and so it cannot detect that they disagree. A definition based on it would assign a "slope" of 0 to a corner, where the graph has no tangent line. The quotient in the definition of the derivative demands agreement from both sides.