SME
Calculus I · Problem type
The chain rule on a power
Use this for “Find f′(1) for f(x) = (2x + 1)³.”
Chart
f(1) = 27
Tangent at x = 1: slope f′(1) = 54, y = 54x − 27
This is an example. Type your own values to start a new problem.
f(x)
=
(
x
+
)
f′(
x
)
=
f′(x)
u
Inside value ax + b
calculated
f(x)
Value there
calculated
Formulas
u
=
a
×
x
+
b
3 = 2 × 1 + 1
f
(
x
)
=
u
n
27
=
3
3
f
′(
x
)
=
n
×
u
n
− 1
×
a
54
=
3
×
3
3 − 1
×
2
Assumptions
•
Chain rule: when f is an outside function of u and u is an inside function of x, f′(x) = (outside slope at u) × (inside slope).
•
Here the inside is u = ax + b, with slope a, and the outside is uⁿ, with slope n·uⁿ⁻¹ by the power rule.
•
So f′(x) = n(ax + b)ⁿ⁻¹ × a. Leaving off the × a is the usual slip.
•
n is a whole number 1–8 and a ≠ 0. The graph meets the x-axis only where the inside is 0, at x = −b ÷ a.
Step-by-step
WE KNOW
a = 2, b = 1, n = 3, x = 1
FIND
inside value ax + b (u), value there (f(x)), slope of the tangent (f′(x))
Step 1 · Find inside value ax + b (u)
Use u = a × x + b
Work
out
the
inside
first:
a
times
x
,
plus
b
.
u
=
2
×
1
+
1
u
=
3
Step 2 · Find value there (f(x))
Use f(x) = uⁿ
Raise
the
inside
value
to
the
power
n
.
f
(
x
)
=
3
3
f
(
x
)
=
27
Step 3 · Find slope of the tangent (f′(x))
Use f′(x) = n × uⁿ⁻¹ × a
Chain
rule:
the
outside’s
slope
n
·uⁿ⁻¹
at
the
inside
value,
times
the
inside’s
slope
a
.
f
′(
x
)
=
3
×
3
3 − 1
×
2
f
′(
x
)
=
3
×
3
2
×
2
f
′(
x
)
=
3
×
9
×
2
f
′(
x
)
=
27
×
2
f
′(
x
)
=
54
Check
Put the numbers back in. Both sides should match.
3 = 2 × 1 + 1 ✓
27
=
3
3
✓
54
=
3
×
3
3 − 1
×
2
✓
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