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AP Calculus BC
Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
3.1 The Chain Rule with Tables and Graphs
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What does the Chain Rule state for composite functions?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
The Chain Rule is used when differentiating a function composed of two or more simpler
functions
.
Steps to apply the Chain Rule for
y
=
y =
y
=
(
3
x
2
+
(3x^{2} +
(
3
x
2
+
5
)
3
5)^{3}
5
)
3
1️⃣ Identify outer function
f
(
u
)
=
f(u) =
f
(
u
)
=
u
3
u^{3}
u
3
2️⃣ Identify inner function
g
(
x
)
=
g(x) =
g
(
x
)
=
3
x
2
+
3x^{2} +
3
x
2
+
5
5
5
3️⃣ Find derivatives
f
′
(
u
)
=
f'(u) =
f
′
(
u
)
=
3
u
2
3u^{2}
3
u
2
and
g
′
(
x
)
=
g'(x) =
g
′
(
x
)
=
6
x
6x
6
x
4️⃣ Apply the Chain Rule:
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
3
(
3
x
2
+
3(3x^{2} +
3
(
3
x
2
+
5
)
2
⋅
6
x
5)^{2} \cdot 6x
5
)
2
⋅
6
x
The Chain Rule is used to differentiate composite
functions
What is the Chain Rule formula for
y
=
y =
y
=
f
(
g
(
x
)
)
f(g(x))
f
(
g
(
x
))
?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
The Chain Rule is used when differentiating functions raised to a
power
.
What is the derivative of
y
=
y =
y
=
(
3
x
2
+
(3x^{2} +
(
3
x
2
+
5
)
3
5)^{3}
5
)
3
?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
18
x
(
3
x
2
+
18x(3x^{2} +
18
x
(
3
x
2
+
5
)
2
5)^{2}
5
)
2
What is a composite function formed by nesting one function inside another?
f
(
g
(
x
)
)
f(g(x))
f
(
g
(
x
))
In a composite function, the function evaluated first is called the
inner
function.
If
f
(
x
)
=
f(x) =
f
(
x
)
=
x
2
x^{2}
x
2
and
g
(
x
)
=
g(x) =
g
(
x
)
=
2
x
+
2x +
2
x
+
3
3
3
, what is f(g(x))</latex>?
(
2
x
+
3
)
2
(2x + 3)^{2}
(
2
x
+
3
)
2
Steps to apply the Chain Rule using tables
1️⃣ Identify outer and inner functions
2️⃣ Find derivative values from tables
3️⃣ Apply the Chain Rule formula:
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
4️⃣ Substitute values into the formula
Given tables for
f
(
x
)
f(x)
f
(
x
)
and
g
(
x
)
g(x)
g
(
x
)
, find
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at x = 2</latex> where
y
=
y =
y
=
f
(
g
(
x
)
)
f(g(x))
f
(
g
(
x
))
and
g
(
2
)
=
g(2) =
g
(
2
)
=
3
3
3
,
g
′
(
2
)
=
g'(2) =
g
′
(
2
)
=
−
1
- 1
−
1
,
f
′
(
3
)
=
f'(3) =
f
′
(
3
)
=
4
4
4
.
−
4
- 4
−
4
When applying the Chain Rule, the derivative of the inner function is multiplied by the derivative of the
outer
function.
What is the derivative of
y
=
y =
y
=
(
2
x
+
1
)
2
(2x + 1)^{2}
(
2
x
+
1
)
2
using the Chain Rule?
4
(
2
x
+
1
)
4(2x + 1)
4
(
2
x
+
1
)
A composite function is created by evaluating one function at the
output
of another.
What is the formula for the Chain Rule?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
The Chain Rule involves differentiating the outer function evaluated at the inner function and multiplying by the derivative of the
inner
function.
If
f
(
u
)
=
f(u) =
f
(
u
)
=
u
2
u^{2}
u
2
and g(x) = 2x + 1</latex>, then
y
=
y =
y
=
(
2
x
+
1
)
2
(2x + 1)^{2}
(
2
x
+
1
)
2
.
What is the derivative of
f
(
u
)
=
f(u) =
f
(
u
)
=
u
2
u^{2}
u
2
?
f
′
(
u
)
=
f'(u) =
f
′
(
u
)
=
2
u
2u
2
u
The derivative of
g
(
x
)
=
g(x) =
g
(
x
)
=
2
x
+
2x +
2
x
+
1
1
1
is 2.
The derivative of y = (2x + 1)^{2}</latex> is
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
4
(
2
x
+
1
)
4(2x + 1)
4
(
2
x
+
1
)
.
What is a composite function formally defined as?
f
(
g
(
x
)
)
f(g(x))
f
(
g
(
x
))
Match the key components of a composite function with their definitions:
Inner function ↔️
g
(
x
)
g(x)
g
(
x
)
Outer function ↔️
f
(
u
)
f(u)
f
(
u
)
, where
u
=
u =
u
=
g
(
x
)
g(x)
g
(
x
)
If
f
(
x
)
=
f(x) =
f
(
x
)
=
x
2
x^{2}
x
2
and
g
(
x
)
=
g(x) =
g
(
x
)
=
2
x
+
2x +
2
x
+
3
3
3
, then f(g(x)) = (2x + 3)^{2}</latex> is the composite function.
When using tables to apply the Chain Rule, you must identify the outer and
inner
functions.
What is the Chain Rule formula when using tables?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
At
x
=
x =
x
=
2
2
2
, the value of
d
y
d
x
\frac{dy}{dx}
d
x
d
y
for
f
(
g
(
x
)
)
f(g(x))
f
(
g
(
x
))
is -4.
To apply the Chain Rule using graphs, you must interpret the graphs of
f
′
(
x
)
f'(x)
f
′
(
x
)
and
g
′
(
x
)
g'(x)
g
′
(
x
)
.
What is the second step in applying the Chain Rule using graphs?
Find
g
(
x
)
g(x)
g
(
x
)
at a specific
x
x
x
If
g
(
2
)
=
g(2) =
g
(
2
)
=
3
3
3
and
g
′
(
2
)
=
g'(2) =
g
′
(
2
)
=
−
1
- 1
−
1
, then
f
′
(
3
)
=
f'(3) =
f
′
(
3
)
=
4
4
4
.
The value of
d
y
d
x
\frac{dy}{dx}
d
x
d
y
at
x
=
x =
x
=
2
2
2
when using graphs is -4.
What is the formula for the Chain Rule?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
f
′
(
g
(
x
)
)
⋅
g
′
(
x
)
f'(g(x)) \cdot g'(x)
f
′
(
g
(
x
))
⋅
g
′
(
x
)
To apply the Chain Rule using graphs, you first use the graphs to find
g
(
x
)
g(x)
g
(
x
)
and
g
′
(
x
)
g'(x)
g
′
(
x
)
at a specific x
The graph of
f
′
f'
f
′
is used to find f'(g(x))</latex> in the Chain Rule.
Steps to apply the Chain Rule using graphs:
1️⃣ Identify the outer and inner functions in the composite function
2️⃣ Use the graphs to find
g
(
x
)
g(x)
g
(
x
)
and
g
′
(
x
)
g'(x)
g
′
(
x
)
at a specific
x
x
x
3️⃣ Find
f
′
(
g
(
x
)
)
f'(g(x))
f
′
(
g
(
x
))
using the graph of
f
′
f'
f
′
4️⃣ Apply the Chain Rule formula
What values do you need to find from the graph of
g
g
g
to apply the Chain Rule at
x
=
x =
x
=
2
2
2
?
g
(
2
)
g(2)
g
(
2
)
and
g
′
(
2
)
g'(2)
g
′
(
2
)
To find
f
′
(
g
(
x
)
)
f'(g(x))
f
′
(
g
(
x
))
, you use the graph of
f
′
f'
f
′
and evaluate it at g(x)
The Chain Rule requires you to find both
g
(
x
)
g(x)
g
(
x
)
and
g
′
(
x
)
g'(x)
g
′
(
x
)
using the graph of
g
g
g
.
What does the Chain Rule allow you to differentiate?
Composite functions
Match the step with its description in applying the Chain Rule using graphs:
Identify outer and inner functions ↔️ Determine the functions in the composite function
Use graphs to find
g
(
x
)
g(x)
g
(
x
)
and
g
′
(
x
)
g'(x)
g
′
(
x
)
↔️ Evaluate the inner function and its derivative
Find
f
′
(
g
(
x
)
)
f'(g(x))
f
′
(
g
(
x
))
using the graph of
f
′
f'
f
′
↔️ Evaluate the derivative of the outer function at
g
(
x
)
g(x)
g
(
x
)
Apply the Chain Rule ↔️ Multiply
f
′
(
g
(
x
)
)
f'(g(x))
f
′
(
g
(
x
))
and
g
′
(
x
)
g'(x)
g
′
(
x
)
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