MechForge · Calculus · Chapter 2: Limits and Derivatives

2.1

The Tangent and Velocity Problems

Two classic problems motivate the entire subject. Finding the slope of a line tangent to a curve and finding the instantaneous velocity of a moving object both require the same idea: a quantity approached by a sequence of approximations. That idea is the limit.

The Tangent Problem

Tangent line

A line that touches a curve at a point and has the same direction as the curve there. For a circle it meets the curve exactly once, but for general curves a one-point rule is inadequate, so the tangent is instead defined as a limit of secant lines.

Secant line

A line through two points of a curve. As the second point slides toward the point of tangency, the secant slopes approach the slope of the tangent line.

Slope of a secant line

\[ m_{PQ} = \frac{f(x) - f(a)}{x - a} \]

Slope of the tangent as a limit of secant slopes

\[ m = \lim_{Q \to P} m_{PQ} = \lim_{x \to a}\frac{f(x)-f(a)}{x-a} \]

Figure 2.1.1As point Q slides toward P along the curve, the secant lines PQ pivot toward the tangent line t. (Original redrawn diagram.)

Example 1 Tangent to a parabola

Find an equation of the tangent line to the parabola \( y = x^2 \) at the point \( P(1,1) \).

Solution

With one known point we approximate the slope using a nearby point \( Q(x, x^2) \): \( \displaystyle m_{PQ} = \frac{x^2 - 1}{x - 1} \).

For \( Q(1.5,\,2.25) \): \( m_{PQ} = \dfrac{2.25-1}{1.5-1} = \dfrac{1.25}{0.5} = 2.5 \). Closer values of \(x\) give slopes approaching 2.

Take the limit: \( \displaystyle m = \lim_{x\to 1}\frac{x^2-1}{x-1} = 2 \).

Point-slope form gives \( y - 1 = 2(x-1) \), i.e. \( y = 2x - 1 \).

Example 2 Slope from experimental data

A capacitor discharges through a flash unit. Given the charge \(Q\) (microcoulombs) remaining at time \(t\) (seconds), estimate the slope of the tangent at \( t = 0.04 \), which represents the current.

Solution

Plot the data and compute secant slopes between the point at \(t=0.04\) and neighbouring data points.

The slopes of the two closest secant lines bracket the answer; averaging them gives roughly \(-675\) microamperes.

A second estimate, by measuring the sides of a triangle on a drawn tangent, gives about \(-670\) — consistent with the data method.

The Velocity Problem

Average velocity

Change in position divided by elapsed time over an interval. It equals the slope of the secant line on the position-versus-time graph.

Instantaneous velocity

The limiting value of average velocities as the time interval shrinks to zero. It equals the slope of the tangent line to the position graph at that instant.

Average velocity over a short interval

\[ \text{average velocity} = \frac{\text{change in position}}{\text{time elapsed}} = \frac{s(a+h)-s(a)}{h} \]

Example 3 A falling ball

A ball is dropped from a 450 m tower. Using Galileo's law \( s(t) = 4.9t^2 \), find the velocity after 5 seconds.

Solution

Average velocity from \(t=5\) to \(t=5.1\): \( \dfrac{4.9(5.1)^2 - 4.9(5)^2}{0.1} = 49.49 \) m/s.

Shrinking the interval (to \([5,5.05]\), \([5,5.01]\), …) produces average velocities approaching 49 m/s.

The instantaneous velocity at \(t=5\) is the limit: \( v = 49 \) m/s.

Instructor Note

Both problems reduce to the same limit, \( \lim_{h\to 0}\dfrac{f(a+h)-f(a)}{h} \). Once you internalize that "slope of tangent" and "instantaneous velocity" are the same calculation, the rest of the chapter becomes one idea applied repeatedly. In mechanics this limit reappears as velocity, acceleration, and every rate you'll meet in dynamics — keep it front of mind.

2.2

The Limit of a Function

Having seen where limits arise, we now study them directly using numerical tables, graphs, and an intuitive definition.

Intuitive meaning of a limit

We write \( \lim_{x\to a} f(x) = L \) if the values of \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently close to \(a\) (on either side) but not equal to \(a\). The value of \(f\) at \(a\) itself is irrelevant — only the behaviour near \(a\) matters.

Limit notation

\[ \lim_{x \to a} f(x) = L \qquad\text{equivalently}\qquad f(x) \to L \ \text{as}\ x \to a \]

Example 1 A removable hole

Guess the value of \( \displaystyle \lim_{x\to 1}\frac{x-1}{x^2-1} \).

Solution

The function is undefined at \(x=1\), but the limit ignores that point.

Tabulating values from both sides (e.g. 0.99 → 0.502513, 1.01 → 0.497512) shows them closing in on 0.5.

Therefore \( \lim_{x\to 1}\dfrac{x-1}{x^2-1} = 0.5 \). Reassigning the value at \(x=1\) does not change the limit.

Example 4 A limit that does not exist

Investigate \( \displaystyle \lim_{x\to 0}\sin\frac{\pi}{x} \).

Solution

Sampling at \(x = 1, \tfrac12, \tfrac13, 0.1, 0.01, \dots\) all give 0, tempting a guess of 0.

But \(f(x)=1\) for infinitely many \(x\) near 0 as well; the function oscillates between \(-1\) and \(1\) ever faster.

Since the values never settle, \( \lim_{x\to 0}\sin(\pi/x) \) does not exist. Sampling alone can mislead.

One-Sided Limits

Left-hand and right-hand limits

\( \lim_{x\to a^-} f(x) = L \) means \(f(x)\) approaches \(L\) as \(x\) approaches \(a\) from values less than \(a\); \( \lim_{x\to a^+} f(x) = L \) uses values greater than \(a\).

Theorem

Two-sided limit via one-sided limits

\( \displaystyle \lim_{x\to a} f(x) = L \) if and only if both one-sided limits exist and equal \(L\):

\[ \lim_{x\to a} f(x) = L \iff \lim_{x\to a^-} f(x) = L = \lim_{x\to a^+} f(x). \]

Example 6 The Heaviside function

For \( H(t) = 0 \) if \(t<0\) and \( H(t)=1 \) if \(t\ge 0\), examine the limit at 0.

Solution

From the left, \(H(t)\to 0\); from the right, \(H(t)\to 1\).

The one-sided limits differ, so \( \lim_{t\to 0} H(t) \) does not exist. (This models a current switched on at \(t=0\).)

Infinite Limits

Infinite limit

\( \lim_{x\to a} f(x) = \infty \) means \(f(x)\) can be made arbitrarily large by taking \(x\) close enough to \(a\). The symbol \(\infty\) is not a number; the limit does not "exist" in the usual sense — the notation just describes how it fails.

Vertical asymptote

The line \(x=a\) is a vertical asymptote of \(y=f(x)\) if at least one one-sided limit of \(f\) at \(a\) is \(+\infty\) or \(-\infty\).

Figure 2.2.1Both branches of \(y = 1/x^2\) rise without bound near \(x=0\), so the \(y\)-axis is a vertical asymptote. (Original redrawn diagram.)

Example 10 Asymptotes of the tangent function

Find the vertical asymptotes of \( f(x) = \tan x \).

Solution

Write \( \tan x = \dfrac{\sin x}{\cos x} \); asymptotes are possible where \( \cos x = 0 \).

Near \( x = \pi/2 \), \( \cos x \to 0 \) while \( \sin x \) stays near 1, so \( \tan x \to +\infty \) from the left and \( -\infty \) from the right.

The lines \( x = (2n+1)\pi/2 \) for every integer \(n\) are all vertical asymptotes.

2.3

Calculating Limits Using the Limit Laws

Tables and graphs can mislead. The Limit Laws let us compute limits exactly from a few basic principles.

Law

The Limit Laws

If \(c\) is a constant and \( \lim_{x\to a} f(x) \) and \( \lim_{x\to a} g(x) \) exist, then:

Sum & Difference Laws

\[ \lim_{x\to a}\big[f(x)\pm g(x)\big] = \lim_{x\to a} f(x) \pm \lim_{x\to a} g(x) \]

Constant Multiple & Product Laws

\[ \lim_{x\to a}\big[c\,f(x)\big] = c\lim_{x\to a} f(x), \qquad \lim_{x\to a}\big[f(x)g(x)\big] = \lim_{x\to a} f(x)\cdot \lim_{x\to a} g(x) \]

Quotient Law

\[ \lim_{x\to a}\frac{f(x)}{g(x)} = \frac{\lim_{x\to a} f(x)}{\lim_{x\to a} g(x)} \qquad \text{if } \lim_{x\to a} g(x) \neq 0 \]

Power & Root Laws

\[ \lim_{x\to a}\big[f(x)\big]^n = \Big[\lim_{x\to a} f(x)\Big]^n, \qquad \lim_{x\to a}\sqrt[n]{f(x)} = \sqrt[n]{\lim_{x\to a} f(x)} \]

Direct Substitution Property

If \(f\) is a polynomial or a rational function and \(a\) is in the domain of \(f\), then \( \lim_{x\to a} f(x) = f(a) \). Such functions are exactly the ones that are continuous at \(a\) (Section 2.5).

Example 3 A 0/0 form solved by factoring

Find \( \displaystyle \lim_{x\to 1}\frac{x^2-1}{x-1} \).

Solution

Direct substitution fails: both numerator and denominator are 0 at \(x=1\).

Factor: \( \dfrac{x^2-1}{x-1} = \dfrac{(x-1)(x+1)}{x-1} = x+1 \) for \(x\neq 1\).

Since the limit ignores \(x=1\), \( \lim_{x\to 1}(x+1) = 2 \).

Example 6 Rationalizing the numerator

Find \( \displaystyle \lim_{t\to 0}\frac{\sqrt{t^2+9}-3}{t^2} \).

Solution

Multiply by the conjugate \( \dfrac{\sqrt{t^2+9}+3}{\sqrt{t^2+9}+3} \) to clear the root.

The numerator becomes \( (t^2+9)-9 = t^2 \), cancelling the denominator: \( \dfrac{1}{\sqrt{t^2+9}+3} \).

Now substitute: \( \dfrac{1}{\sqrt{9}+3} = \dfrac{1}{6} \).

Theorem

Order-preserving property

If \( f(x) \le g(x) \) near \(a\) (except possibly at \(a\)) and both limits exist, then \( \lim_{x\to a} f(x) \le \lim_{x\to a} g(x) \).

Theorem

The Squeeze Theorem

If \( f(x) \le g(x) \le h(x) \) near \(a\) (except possibly at \(a\)) and \( \lim_{x\to a} f(x) = \lim_{x\to a} h(x) = L \), then \( \lim_{x\to a} g(x) = L \).

Intuitively, a function trapped between two others that meet at \(L\) is forced to \(L\) as well.

Example 11 Squeeze theorem in action

Show that \( \displaystyle \lim_{x\to 0} x^2\sin\frac{1}{x} = 0 \).

Solution

The Product Law fails because \( \lim_{x\to 0}\sin(1/x) \) does not exist.

Since \( -1 \le \sin(1/x) \le 1 \), multiply by \(x^2 \ge 0\): \( -x^2 \le x^2\sin(1/x) \le x^2 \).

Both bounds \( \pm x^2 \to 0 \), so by the Squeeze Theorem the middle limit is 0.

Instructor Note

When direct substitution gives \(0/0\), that's an indeterminate form — not an answer, but a signal to do algebra first: factor, rationalize, or simplify a complex fraction, then substitute. The Squeeze Theorem is your tool when a function won't factor but you can bound it. Reach for it whenever you see a bounded factor (like sine or cosine) multiplied by something going to zero.

2.4

The Precise Definition of a Limit

Phrases like "close to" are too vague for proofs. The epsilon–delta definition makes the intuitive notion of a limit rigorous.

The ε–δ definition of a limit

\( \lim_{x\to a} f(x) = L \) means: for every \( \varepsilon > 0 \) there exists a \( \delta > 0 \) such that whenever \( 0 < |x-a| < \delta \), it follows that \( |f(x)-L| < \varepsilon \). Read as a challenge — for any error tolerance \(\varepsilon\), you can find a closeness \(\delta\) that guarantees it.

Precise definition of a limit

\[ \forall\, \varepsilon > 0 \ \exists\, \delta > 0 \ : \ 0 < |x-a| < \delta \implies |f(x)-L| < \varepsilon \]

One-sided ε–δ definitions

For the left-hand limit, restrict \(x\) to \( a-\delta < x < a \); for the right-hand limit, restrict \(x\) to \( a < x < a+\delta \). Everything else is identical.

Figure 2.4.1Whenever \(x\) lies within \(\delta\) of \(a\), the curve stays inside the horizontal \(\varepsilon\)-band around \(L\). (Original redrawn diagram.)

Example 2 A linear ε–δ proof

Prove that \( \displaystyle \lim_{x\to 3}(4x-5) = 7 \).

Solution

Guess \(\delta\): \( |(4x-5)-7| = |4x-12| = 4|x-3| \). To make this \(<\varepsilon\), require \( |x-3| < \varepsilon/4 \).

Choose \( \delta = \varepsilon/4 \).

Verify: if \( 0 < |x-3| < \delta \), then \( |(4x-5)-7| = 4|x-3| < 4\delta = 4(\varepsilon/4) = \varepsilon \). ∎

Precise definition of an infinite limit

\( \lim_{x\to a} f(x) = \infty \) means: for every \( M > 0 \) there is a \( \delta > 0 \) such that \( 0 < |x-a| < \delta \) implies \( f(x) > M \). A negative-infinity version replaces \( f(x) > M \) with \( f(x) < N \) for every \( N < 0 \).

Proof — Sum Law

The Sum Law follows from the definition

If \( \lim_{x\to a} f(x) = L \) and \( \lim_{x\to a} g(x) = M \), then \( \lim_{x\to a}[f(x)+g(x)] = L+M \). The proof picks \(\delta_1,\delta_2\) making each of \( |f-L| \) and \( |g-M| \) less than \( \varepsilon/2 \), then uses the Triangle Inequality \( |f+g-(L+M)| \le |f-L| + |g-M| < \varepsilon \).

2.5

Continuity

The functions whose limits equal their values are called continuous — their graphs have no breaks.

Continuity at a number

A function \(f\) is continuous at \(a\) if \( \lim_{x\to a} f(x) = f(a) \). This single equation silently requires three things: \(f(a)\) is defined, the limit exists, and the two are equal.

Definition of continuity at a point

\[ \lim_{x \to a} f(x) = f(a) \]

Types of discontinuity

Removable: the limit exists but doesn't equal \(f(a)\) (a hole), fixable by redefining one point. Jump: the one-sided limits exist but differ. Infinite: the function blows up, as at a vertical asymptote.

One-sided continuity & continuity on an interval

\(f\) is continuous from the right at \(a\) if \( \lim_{x\to a^+} f(x) = f(a) \), and from the left if \( \lim_{x\to a^-} f(x) = f(a) \). It is continuous on an interval if it is continuous at every number there (using one-sided continuity at endpoints).

Figure 2.5.1A removable discontinuity (open hole) and a jump discontinuity. (Original redrawn diagram.)

Theorem

Combining continuous functions

If \(f\) and \(g\) are continuous at \(a\) and \(c\) is a constant, then \( f+g,\ f-g,\ cf,\ fg \), and \( f/g \) (where \( g(a)\neq 0 \)) are all continuous at \(a\). A composition of continuous functions is continuous: if \(g\) is continuous at \(a\) and \(f\) is continuous at \(g(a)\), then \( f\circ g \) is continuous at \(a\).

Theorem

Continuity of familiar functions

Polynomials, rational functions, root functions, trigonometric and inverse trigonometric functions, exponential functions, and logarithmic functions are all continuous at every number in their domains.

Example 5 Limit by continuity

Find \( \displaystyle \lim_{x\to -2}\frac{x^3+2x^2-1}{5-3x} \).

Solution

The function is rational and \(x=-2\) is in its domain (denominator \(\neq 0\)), so it is continuous there.

Continuity lets us substitute directly: \( \dfrac{(-2)^3+2(-2)^2-1}{5-3(-2)} = -\dfrac{1}{11} \).

Theorem

The Intermediate Value Theorem

If \(f\) is continuous on the closed interval \([a,b]\) and \(N\) is any value between \(f(a)\) and \(f(b)\) with \( f(a)\neq f(b) \), then there exists \(c\) in \((a,b)\) with \( f(c) = N \). A continuous graph cannot jump over a horizontal line — it must cross it.

Example 10 Locating a root

Show that \( 4x^3-6x^2+3x-2 = 0 \) has a root between 1 and 2.

Solution

Let \( f(x) = 4x^3-6x^2+3x-2 \). Then \( f(1) = -1 < 0 \) and \( f(2) = 12 > 0 \).

Since \(N=0\) lies between \(f(1)\) and \(f(2)\) and \(f\) is continuous, the IVT guarantees some \(c\) in \((1,2)\) with \( f(c)=0 \).

Reapplying the IVT narrows the root: a sign change between \(f(1.22)<0\) and \(f(1.23)>0\) places it in \((1.22,\,1.23)\).

Instructor Note

The Intermediate Value Theorem is the workhorse behind numerical root-finding (bisection method). In engineering you'll use exactly this sign-change logic to locate where a stress, temperature, or equilibrium condition crosses a critical value. The theorem only promises a root exists — it doesn't hand you the value, which is why iterative methods matter.

2.6

Limits at Infinity; Horizontal Asymptotes

Instead of letting \(x\) approach a finite number, we now let \(x\) grow without bound and ask what happens to \(f(x)\).

Limit at infinity

\( \lim_{x\to\infty} f(x) = L \) means \(f(x)\) can be made arbitrarily close to \(L\) by taking \(x\) sufficiently large. A parallel definition uses \( x\to -\infty \) for large negative \(x\).

Horizontal asymptote

The line \(y=L\) is a horizontal asymptote of \(y=f(x)\) if either \( \lim_{x\to\infty} f(x) = L \) or \( \lim_{x\to -\infty} f(x) = L \). A curve may have two different horizontal asymptotes (e.g. \( y=\tan^{-1}x \) with \( \pm\pi/2 \)).

Theorem

Reciprocal power rule at infinity

If \( r > 0 \) is rational, then \( \displaystyle \lim_{x\to\infty}\frac{1}{x^r} = 0 \), and (where \(x^r\) is defined for all \(x\)) \( \displaystyle \lim_{x\to -\infty}\frac{1}{x^r} = 0 \).

Key reciprocal limit

\[ \lim_{x \to \infty}\frac{1}{x^r} = 0 \qquad (r > 0) \]

Example 3 Limit of a rational function

Evaluate \( \displaystyle \lim_{x\to\infty}\frac{3x^2-x-2}{5x^2+4x+1} \).

Solution

Divide numerator and denominator by the highest power in the denominator, \(x^2\).

This gives \( \dfrac{3 - 1/x - 2/x^2}{5 + 4/x + 1/x^2} \).

Every reciprocal-power term \(\to 0\), leaving \( \dfrac{3-0-0}{5+0+0} = \dfrac{3}{5} \). So \(y=3/5\) is a horizontal asymptote.

Example 5 A difference of roots

Compute \( \displaystyle \lim_{x\to\infty}\left(\sqrt{x^2+1}-x\right) \).

Solution

This is an \(\infty - \infty\) form; multiply by the conjugate \( \dfrac{\sqrt{x^2+1}+x}{\sqrt{x^2+1}+x} \).

The numerator simplifies to \( (x^2+1)-x^2 = 1 \), giving \( \dfrac{1}{\sqrt{x^2+1}+x} \).

As \(x\to\infty\) the denominator grows without bound, so the limit is 0.

Infinite limits at infinity

\( \lim_{x\to\infty} f(x) = \infty \) records that \(f(x)\) grows without bound as \(x\) does. For example \( \lim_{x\to\infty} x^3 = \infty \) while \( \lim_{x\to -\infty} x^3 = -\infty \). Exponential growth eventually outpaces any power: \( e^x \) dwarfs \( x^3 \) for large \(x\).

2.7

Derivatives and Rates of Change

The tangent-slope limit and the velocity limit are the same object — the derivative — and it measures a rate of change in any science.

Tangent line via the derivative

The tangent line to \(y=f(x)\) at \(P(a,f(a))\) is the line through \(P\) whose slope is the limit of secant slopes, provided that limit exists.

Slope of the tangent line (two forms)

\[ m = \lim_{x\to a}\frac{f(x)-f(a)}{x-a} = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \]

Instantaneous velocity

For a position function \( s=f(t) \), the velocity at \( t=a \) is the limit of average velocities over shrinking intervals — exactly the derivative of \(f\) at \(a\).

Instantaneous velocity

\[ v(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \]

The derivative at a number

The derivative of \(f\) at \(a\), written \( f'(a) \), is the limit defining the tangent slope and the velocity. Because the same limit measures rate of reaction, marginal cost, power, and population growth, it earns its own name and notation.

Definition of the derivative at a number

\[ f'(a) = \lim_{h\to 0}\frac{f(a+h)-f(a)}{h} \]

Figure 2.7.1The difference quotient is the rise \(f(a+h)-f(a)\) over the run \(h\); its limit as \(h\to 0\) is the derivative. (Original redrawn diagram.)

Example 4 Derivative of a quadratic

Find the derivative of \( f(x) = x^2 - 8x + 9 \) at a general number \(a\).

Solution

Form the difference quotient: \( \dfrac{[(a+h)^2 - 8(a+h)+9] - [a^2-8a+9]}{h} \).

Expand and simplify the numerator to \( 2ah + h^2 - 8h \), then divide by \(h\): \( 2a + h - 8 \).

Take the limit \(h\to 0\): \( f'(a) = 2a - 8 \).

Average vs. instantaneous rate of change

The average rate of change of \(y=f(x)\) over \([x_1,x_2]\) is \( \Delta y/\Delta x \), the secant slope. Its limit as \( \Delta x \to 0 \) is the instantaneous rate of change — the derivative \( f'(x_1) \), the tangent slope.

Instantaneous rate of change

\[ \text{rate} = \lim_{\Delta x \to 0}\frac{\Delta y}{\Delta x} = \lim_{x_2 \to x_1}\frac{f(x_2)-f(x_1)}{x_2 - x_1} \]

Instructor Note

"Derivative = instantaneous rate of change" is the sentence to memorize. Velocity is the rate of change of position; acceleration is the rate of change of velocity; power is the rate of change of work; marginal cost is the rate of change of cost. Same limit, different units. When you read a derivative's units (e.g. dollars per yard, m/s), you immediately know what it physically means.

2.8

The Derivative as a Function

By letting the base point vary, the derivative becomes a new function in its own right.

The derivative as a function

Replacing the fixed number \(a\) with a variable \(x\) produces the function \(f'\), whose value at each \(x\) is the slope of the tangent to \(y=f(x)\) there. Its domain is the set of \(x\) where the limit exists, which may be smaller than the domain of \(f\).

The derivative function

\[ f'(x) = \lim_{h\to 0}\frac{f(x+h)-f(x)}{h} \]

Leibniz and other notations

For \(y=f(x)\), the derivative is written \( f'(x) = y' = \dfrac{dy}{dx} = \dfrac{df}{dx} = \dfrac{d}{dx}f(x) = Df(x) \). The symbol \( dy/dx \) is a single object, not a literal quotient (for now), though it behaves suggestively.

Differentiable

\(f\) is differentiable at \(a\) if \( f'(a) \) exists, and differentiable on an interval if it is differentiable at every point there.

Example 3 Derivative of the square root

If \( f(x) = \sqrt{x} \), find \( f'(x) \) and state its domain.

Solution

Form \( \dfrac{\sqrt{x+h}-\sqrt{x}}{h} \) and rationalize the numerator by multiplying by \( \dfrac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+\sqrt{x}} \).

The numerator becomes \( (x+h)-x = h \), cancelling to \( \dfrac{1}{\sqrt{x+h}+\sqrt{x}} \).

Take \(h\to 0\): \( f'(x) = \dfrac{1}{2\sqrt{x}} \), valid for \( x>0 \). The domain shrinks from \([0,\infty)\) to \((0,\infty)\).

Theorem

Differentiability implies continuity

If \(f\) is differentiable at \(a\), then \(f\) is continuous at \(a\). The converse is false: \( f(x) = |x| \) is continuous at 0 but not differentiable there because the graph has a corner.

How a function fails to be differentiable

Three common failures: a corner (left and right slopes differ, as in \(|x|\)), a discontinuity (no derivative since continuity fails), and a vertical tangent (the slope limit is \(\pm\infty\)).

Figure 2.8.1Three ways a function fails to be differentiable at a point: a corner, a discontinuity, and a vertical tangent. (Original redrawn diagram.)

Higher Derivatives

Second derivative and beyond

The derivative of \(f'\) is the second derivative \(f''\), interpreted as the rate of change of the slope. Continuing gives \( f''' \), \( f^{(4)} \), and in general the \(n\)th derivative \( f^{(n)} \).

Second derivative (Leibniz notation)

\[ f''(x) = \frac{d}{dx}\!\left(\frac{dy}{dx}\right) = \frac{d^2 y}{dx^2} \]

Acceleration and jerk

For a position function \( s=s(t) \): the first derivative is velocity \( v = ds/dt \), the second is acceleration \( a = d^2s/dt^2 \), and the third is jerk \( j = d^3s/dt^3 \) — the rate of change of acceleration, felt as an abrupt change in motion.

Velocity, acceleration, jerk

\[ v(t) = s'(t), \qquad a(t) = v'(t) = s''(t), \qquad j(t) = a'(t) = s'''(t) \]

Example 7 Higher derivatives of a cubic

If \( f(x) = x^3 - x \), find \( f'''(x) \) and \( f^{(4)}(x) \).

Solution

From earlier, \( f'(x) = 3x^2 - 1 \) and \( f''(x) = 6x \).

The slope of \( y = 6x \) is constant, so \( f'''(x) = 6 \).

A constant has zero slope, so \( f^{(4)}(x) = 0 \) for all \(x\).

Instructor Note

In dynamics you'll live in this hierarchy: position → velocity → acceleration → jerk. Smooth ride design (elevators, trains, robotics) is largely about controlling jerk, not just acceleration — sudden acceleration changes are what passengers feel as a lurch. Note too that differentiability is stronger than continuity: a continuous signal with a sharp corner has no well-defined instantaneous rate there, which matters when you model real measured data.