SME
Grade 12 Math · Problem type
From a secant slope to the derivative
Use this for “Find the slope of the secant of x² from 3 to 3.1, then the derivative at 3.”
Chart
Secant
slope
f
(3.1) −
f
(3)
0.1
=
6.1
As h → 0 the secant turns into the tangent at P, slope 6
This is an example. Type your own values to start a new problem.
f(x)
=
a
x²
+
b
x
+
c
m
=
f(
+
)
−
f(
)
h
f′
Derivative, the tangent’s slope
worked out (not typed)
Formulas
m
=
f
(
x
+
h
) −
f
(
x
)
h
,
with
f
(
x
)
=
ax
²
+
bx
+
c
6.1
=
f
(3 + 0.1) −
f
(3)
0.1
,
with
f
(
x
)
=
1
x
²
+
0
x
+
0
f′
=
2
×
a
×
x
+
b
6 = 2 × 1 × 3 + 0
Assumptions
•
The secant through (x, f(x)) and (x + h, f(x + h)) has slope (f(x + h) − f(x)) ÷ h.
•
For f(x) = ax² + bx + c that slope is 2ax + b + ah.
•
As h → 0 the secant turns into the tangent, and its slope into the derivative 2ax + b.
Step-by-step
WE KNOW
a = 1, b = 0, c = 0, x = 3, h = 0.1
FIND
secant slope (m), derivative, the tangent’s slope (f′)
Step 1 · Find secant slope (m)
Use m = (f(x + h) − f(x)) ÷ h, with f(x) = ax² + bx + c
Here
f
(
x
)
=
x
2
.
The
secant’s
slope
is
the
rise
from
(
x
,
f
(
x
))
to
(
x
+
h
,
f
(
x
+
h
))
over
the
run
h
.
m
=
((
a
×
(
x
+
h
)
2
+
b
×
(
x
+
h
)
+
c
)
−
(
a
×
x
2
+
b
×
x
+
c
))
÷
h
m
=
((1
×
(3 + 0.1)
2
+
0
×
(3
+
0.1)
+
0)
−
(1
×
3
2
+
0
×
3
+
0))
÷
0.1
f
(3.1)
−
f
(3)
=
9.61
−
9
=
0.61
0.61
0.1
=
6.1
m
=
6.1
Step 2 · Find derivative, the tangent’s slope (f′)
Use f′ = 2 × a × x + b
The
secant
slope
is
2ax
+
b
+
ah;
let
h
shrink
to
0.
f′ = 2 × 1 × 3 + 0
f′ = 6
Check
Put the numbers back in. Both sides should match.
6.1
=
9.61 − 9
0.1
✓
6 = 2 × 1 × 3 + 0 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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Related lessons
Limits, continuity and an introduction to derivatives
Main lesson
One-sided limits and continuity
Use this for “f(x) = x + 1 for x < 2 and 2x − 3 for x ≥ 2. Is f continuous at 2?”
Limits at infinity
Use this for “Find the limit of (2x + 1) ÷ (x − 3) as x → ∞.” Other powers of x are on “Limits at infinity by degree.”