Interpretation — instantaneous rate of change, slope of tangent: Worked examples

This page is a drill hall. The parent note built the machinery (secant → limit → tangent). Here we throw every kind of input at that machinery so no exam question can surprise you. We compute everything from the difference quotient

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

and never skip the "why". Read the scenario matrix first, then hunt each cell.

The scenario matrix

Every derivative-at-a-point question is one of these case classes. The last column tells you which worked example below nails it.

#	Case class	What makes it tricky	Example
A	Positive slope (curve rising)	baseline sanity check	Ex 1
B	Negative slope (curve falling)	sign of the answer must come out $< 0$	Ex 2
C	Zero slope (flat tangent, a peak/valley)	answer is exactly $0$	Ex 3
D	Denominator function $1/ x$	the $h$ hides inside a fraction	Ex 4
E	Root function $x$	must rationalise, not just expand	Ex 5
F	Word / rate problem with units	interpret sign + attach units	Ex 6
G	Degenerate: corner $∣ x ∣$	limit fails, no derivative	Ex 7
H	Degenerate: vertical tangent $x^{1/3}$	limit is $+ \infty$ , no finite slope	Ex 8
I	Exam twist: tangent parallel to a given line	solve $f^{'} (a) = m$ for $a$	Ex 9

Prerequisites you may want open: Average rate of change (the secant slope), Limits — formal definition (the engine), and Velocity and acceleration (for the word problem).

Ex 1 — Case A: positive slope

Worked example Slope of

$f (x) = x^{2} - 2 x$ at $x = 3$ Forecast: the parabola $y = x^{2} - 2 x$ opens up and $x = 3$ is to the right of its lowest point, so the curve is rising. Guess: a positive number.

Step 1. Form the difference quotient. $\frac{f ( 3 + h ) - f ( 3 )}{h} = \frac{[ ( 3 + h ) ^{2} - 2 ( 3 + h ) ] - [ 9 - 6 ]}{h}$ Why this step? This is the honest slope of the secant through two real points — no division by zero yet.

Step 2. Expand the top: $(3 + h)^{2} = 9 + 6 h + h^{2}$ and $- 2 (3 + h) = - 6 - 2 h$ , so the bracket is $9 + 6 h + h^{2} - 6 - 2 h = 3 + 4 h + h^{2}$ . Subtract $f (3) = 3$ : $numerator = 4 h + h^{2} .$ Why this step? We need the $\frac{0}{0}$ trouble ( $h$ in the bottom) to be matched by an $h$ on top so we can cancel.

Step 3. Cancel and take the limit. $\frac{4 h + h ^{2}}{h} = 4 + h (h \neq = 0), lim_{h \to 0} (4 + h) = 4.$ Why this step? Cancelling kills the illegal $0/0$ before we let $h \to 0$ — "Simplify Before Limit".

So $f^{'} (3) = 4$ .

Verify: positive, as forecast. Sanity check with a tiny secant $h = 0.001$ : $4 + 0.001 = 4.001 \approx 4$ . ✓ Tangent line: $y - 3 = 4 (x - 3)$ .

Figure — Interpretation — instantaneous rate of change, slope of tangent

Ex 2 — Case B: negative slope

Ex 3 — Case C: zero slope (flat tangent at a vertex)

Ex 4 — Case D: a reciprocal function

Worked example Slope of

$f (x) = \frac{1}{x}$ at $x = 2$ Forecast: $1/ x$ is a decreasing curve for $x > 0$ (bigger $x$ → smaller output), so we expect a negative slope, small in size.

Step 1. Difference quotient: $\frac{\frac{1}{2 + h} - \frac{1}{2}}{h} .$ Why this step? Standard secant, but the $h$ is now buried inside a fraction, so we can't cancel yet — we must combine first.

Step 2. Combine the top fractions over the common denominator $2 (2 + h)$ : $\frac{1}{2 + h} - \frac{1}{2} = \frac{2 - ( 2 + h )}{2 ( 2 + h )} = \frac{- h}{2 ( 2 + h )} .$ Why this step? This manufactures an $h$ on top to match the $h$ underneath. Combining fractions is the only tool that reaches a buried $h$ .

Step 3. Divide by $h$ (i.e. multiply by $1/ h$ ): $\frac{- h}{2 ( 2 + h )} \cdot \frac{1}{h} = \frac{- 1}{2 ( 2 + h )}, lim_{h \to 0} \frac{- 1}{2 ( 2 + h )} = \frac{- 1}{4} .$ Why this step? Now the $h$ cancels cleanly and plugging $h = 0$ is legal. So $f^{'} (2) = - \frac{1}{4}$ .

Verify: negative and small, as forecast. Tiny secant $h = 0.001$ : $\frac{1/2.001 - 1/2}{0.001} \approx - 0.2499 \approx - 0.25$ . ✓

Ex 5 — Case E: a square root (rationalising trick)

Ex 6 — Case F: a rate word problem with signs and units

Worked example A ball is thrown up; height

$s (t) = 20 t - 5 t^{2}$ metres, $t$ in seconds. Find its velocity at $t = 1$ s and at $t = 3$ s. Forecast: the ball rises then falls. At $t = 1$ s (still climbing) velocity should be positive; by $t = 3$ s (past the top) it should be negative. Units: metres per second.

Step 1 (build once). $\frac{s ( a + h ) - s ( a )}{h}$ with $s (t) = 20 t - 5 t^{2}$ . Why? Velocity is the instantaneous rate of change of position — see Velocity and acceleration.

Step 2 (general slope). Top $= 20 (a + h) - 5 (a + h)^{2} - [20 a - 5 a^{2}]$ . Expand $- 5 (a + h)^{2} = - 5 a^{2} - 10 ah - 5 h^{2}$ : $top = 20 h - 10 ah - 5 h^{2} .$ Divide by $h$ : $20 - 10 a - 5 h$ . Why? Doing it with a general $a$ lets us reuse the work for both times.

Step 3 (limit, then substitute). $h \to 0 lim (20 - 10 a - 5 h) = 20 - 10 a$ . So $v (a) = 20 - 10 a$ .

At $a = 1$ : $v = 20 - 10 = + 10$ m/s.
At $a = 3$ : $v = 20 - 30 = - 10$ m/s.

Verify: signs match the forecast — $+ 10$ climbing, $- 10$ descending. Units: $[m] / [s]$ = m/s. ✓ The peak (where $v = 0$ ) is at $a = 2$ s, exactly between our two times, which is consistent with the sign flip.

Ex 7 — Case G: degenerate corner (no derivative)

Worked example Does

$f (x) = ∣ x - 1∣$ have a slope at $x = 1$ ? Forecast: $∣ x - 1∣$ makes a sharp V with its point at $x = 1$ . A corner has no single tangent — guess: derivative does not exist.

Step 1. Difference quotient at $a = 1$ : since $f (1) = 0$ , $\frac{∣1 + h - 1∣ - 0}{h} = \frac{∣ h ∣}{h} .$ Why this step? We must test whether the limit even exists before claiming a number.

Step 2. Split by the sign of $h$ — this is the "cover all cases" move:

$h > 0$ : $∣ h ∣ = h \Rightarrow \frac{h}{h} = + 1.$
$h < 0$ : $∣ h ∣ = - h \Rightarrow \frac{- h}{h} = - 1.$

Why this step? The absolute value behaves differently on each side, so the one-sided limits can disagree.

Step 3. Right limit $= + 1$ , left limit $= - 1$ . They disagree, so $h \to 0 lim \frac{∣ h ∣}{h}$ does not exist. Therefore $f^{'} (1)$ does not exist.

Verify: matches forecast — a corner. This is the Differentiability and continuity lesson: continuous everywhere, yet not differentiable at the point.

Ex 8 — Case H: degenerate vertical tangent (limit is infinite)

Worked example Does

$f (x) = x^{1/3}$ (the cube root) have a slope at $x = 0$ ? Forecast: the cube-root curve passes through the origin standing almost straight up. A vertical tangent has no finite slope — guess: the limit blows up to $+ \infty$ , so no finite derivative.

Step 1. At $a = 0$ , $f (0) = 0$ , so the difference quotient is $\frac{( 0 + h ) ^{1/3} - 0}{h} = \frac{h ^{1/3}}{h} = h^{1/3 - 1} = h^{- 2/3} = \frac{1}{h ^{2/3}} .$ Why this step? Writing it as a single power exposes the behaviour as $h \to 0$ .

Step 2. As $h \to 0$ (from either side), $h^{2/3} = (h^{1/3})^{2} \geq 0$ shrinks toward $0^{+}$ , so $\frac{1}{h ^{2/3}}$ grows without bound: $lim_{h \to 0} \frac{1}{h ^{2/3}} = + \infty.$ Why this step? A cube root of a negative number is negative, but squaring it makes the denominator positive on both sides — so both one-sided limits go to $+ \infty$ together.

Step 3. The limit is not a finite number, so $f^{'} (0)$ does not exist (the tangent is vertical, slope " $\infty$ ").

Verify: matches forecast. Check the magnitude at $h = 0.001$ : $h^{- 2/3} = 0.00 1^{- 2/3} = 100$ , and at $h = 1 0^{- 6}$ it is $1 0^{4}$ — clearly growing. ✓

Ex 9 — Case I: exam twist (tangent parallel to a given line)

Worked example For

$f (x) = x^{2}$ , at which point is the tangent parallel to the line $y = 8 x - 3$ ? Forecast: parallel lines share a slope. The line has slope $8$ , so we need $f^{'} (a) = 8$ . Since $y = x^{2}$ steepens as $x$ grows, expect a single positive $a$ .

Step 1. Get the general slope of $x^{2}$ . Difference quotient $\frac{( a + h ) ^{2} - a ^{2}}{h} = \frac{2 ah + h ^{2}}{h} = 2 a + h \to 2 a$ . So $f^{'} (a) = 2 a$ . Why this step? We need slope as a function of position so we can set it equal to a target.

Step 2. Set $f^{'} (a) = 8$ : $2 a = 8 \Rightarrow a = 4.$ Why this step? "Parallel" translates directly into "equal slope"; solving for $a$ locates the point.

Step 3. The point is $(4, f (4)) = (4, 16)$ . Tangent line: $y - 16 = 8 (x - 4) \Rightarrow y = 8 x - 16$ . Why this step? Point–slope form finishes the job once we have the point and slope.

Verify: slope $8$ matches the given line's slope $8$ ✓, and the two lines differ ( $- 16 \neq = - 3$ ), so they are parallel, not the same line. ✓

Recall Rapid self-test across the matrix

Which case gives a negative answer, and why? ::: Case B — a falling curve; the minus sign distributes onto the squared term. Which two cases have "derivative does not exist", and how do they differ? ::: Corner (Ex 7, one-sided limits $+ 1$ vs $- 1$ ) and vertical tangent (Ex 8, limit $\to + \infty$ ). For a reciprocal or root, what algebra unblocks the buried $h$ ? ::: Combine fractions (reciprocal) or multiply by the conjugate (root). "Tangent parallel to line $y = m x + c$ " becomes which equation? ::: $f^{'} (a) = m$ , then solve for $a$ .

Connections

Average rate of change — every example started as a secant slope before the limit.
Limits — formal definition — Ex 7 and Ex 8 are really limit-existence questions.
Derivative as a function — Ex 6 and Ex 9 found $f^{'} (a)$ for general $a$ , i.e. $f^{'} (x)$ .
Power Rule — Ex 5 and Ex 9 secretly confirm it.
Differentiability and continuity — Ex 7, 8 show continuity is not enough for a derivative.
Velocity and acceleration — Ex 6 is the physics face of the same limit.