A curve has no single slope — it has a slope at every point. The derivative is the machine that hands you that slope on demand.
Read the graph
What is the slope of the tangent line at (1, 1)?
The derivative of x² is 2x, and at x = 1 that gives 2 — and the picture agrees: the tangent climbs two squares for every one across. A confuses the point's coordinates with its slope; the curve is already climbing faster than 45° at (1, 1).
A straight line has one slope everywhere. A curve is steeper in some places than others — so "the slope of the curve at a point" means the slope of the tangent line there: the straight line that just grazes the curve at that point, matching its direction exactly.
For f(x) = x², the machine is f′(x) = 2x — feed it a point, get the slope there.
f(x) = x² → f′(x) = 2x
by the power rule — next lesson's job
f′(1) = 2
slope of the tangent at (1, 1)
f′(0) = 0, f′(−1) = −2
flat at the bottom, falling on the left
At (1, 1) it is tempting to answer “slope 1” — just reading the point. But f(1) = 1 is how high the curve is; f′(1) = 2(1) = 2 is how steep it is. At x = 3 the gap is wider still: height f(3) = 9, slope f′(3) = 6. Always feed the x into f′, not f.
Two points on the tangent give its slope the same way: a tangent through (1, 5) and (4, −1) has slope (−1 − 5) ÷ (4 − 1) = −2, so f′ = −2 at the point of contact. And a tangent given as an equation hands over both numbers: if the tangent at x = 2 is y = 5x − 4, then f′(2) = 5 (its slope) and f(2) = 5(2) − 4 = 6 (it touches the curve there).
| f′ at the point | The curve is… |
|---|---|
| positive | rising left → right |
| negative | falling left → right |
| zero | flat — a peak, valley or shelf |
On y = x²: negative slopes on the left of the vertex, zero at the bottom, positive on the right. That "f′ = 0 at the flat spot" is the key that unlocks maximum–minimum problems.
For f(x) = x² + 2x, f′(x) = 2x + 2. Where is the tangent parallel to y = 8x − 5? Where is the graph falling?
parallel → slope 8
parallel lines share a slope
2x + 2 = 8 → x = 3
set f′ equal to it — not f
f(3) = 9 + 6 = 15 → point (3, 15)
the height comes from f
falling: 2x + 2 < 0 → x < −1
“falling” means f′ negative
Solving x² + 2x = 8 answers “where is the height 8?” — a different question. And “where is the graph falling” is about the sign of f′, not where f is negative.
A letter in f′ is found the same way. If f′(x) = 2x + c and the slope at x = 2 is 7, then 4 + c = 7, so c = 3, and then f′(0) = 0 + 3 = 3.
A flat tangent, fractions included. f(x) = x² − 5x + 1 with f′(x) = 2x − 5. Where is the tangent horizontal?
2x − 5 = 0 → x = 52
horizontal means slope 0
f(52) = 254 − 504 + 44 = −214
the height comes from f
point (52, −214)
the bottom of this parabola
What is different: the given slope is 0, and the answer is not a whole number. Keep quarters until the end.
Perpendicular, not parallel. Same f. Where is the tangent perpendicular to y = −½x + 4?
slope needed = 2
negative reciprocal of −½
2x − 5 = 2 → x = 72
set f′ equal to it
f(72) = 494 − 704 + 44 = −174 → (72, −174)
check: 2 × (−½) = −1 ✓
What is different: the slope is not read straight off the line; it is flipped and its sign changed.
That gives 2x − 5 = −½, x = 94 — the point where the tangent is PARALLEL to the line. Parallel copies the slope; perpendicular flips it.
Two answers, or none. g(x) = x³ + 2x with g′(x) = 3x² + 2. Where is the slope 29? Where is it 1?
3x² + 2 = 29 → x² = 9
slope 29
x = 3 or −3
a square has two roots
3x² + 2 = 1 → x² = −13
slope 1
no such point
g′ is never below 2
What is different: f′ is a quadratic, so a slope can occur twice or never. This curve is never flat and never falling.
Where is a cubic falling? For h(x) = x³ − 12x, h′(x) = 3x² − 12 < 0 when x² < 4, that is −2 < x < 2 — the stretch between its two flat points.
f′(a) = limit as h → 0 of [f(a + h) − f(a)] ÷ h
The slope between x = a and a point h further along, as h shrinks to nothing. Find it for f(x) = 2x² at a = −1.
f(−1 + h) = 2(1 − 2h + h²) = 2 − 4h + 2h²
expand carefully
f(−1 + h) − f(−1) = −4h + 2h²
f(−1) = 2; the 2s cancel
÷ h: −4 + 2h
every term had an h
h → 0: f′(−1) = −4
falling, steeply
Same idea for our parabola: [(1 + h)² − 1] ÷ h = (2h + h²) ÷ h = 2 + h → 2, the f′(1) from Step 2. When an exam shows this limit, recognise it as “f′(a)” — if you already know f′, just plug a in.
Expect “at what x is the slope m”, “where is the curve rising/falling”, a graph with points to classify by the sign of f′, or the h-limit in disguise. Before answering, ask: is this asking for a height (use f) or a steepness (use f′)?
The problem
The derivative of f(x) = x³ is f′(x) = 3x². What is the slope of the tangent to the curve at x = 1?
Feed the point into the slope machine: f′(1) = 3(1)² = 3. B mistakes the point's height f(1) = 1 for its slope. C uses the lowered exponent alone, 3 − 1 = 2, and forgets the 3 in front. D differentiates twice: 3x² becomes 6x, which is 6 at x = 1. The derivative is evaluated once, never guessed from the picture's steep look.
