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AP Calculus AB
Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
3.2 Implicit Differentiation
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What is the form of an explicit function?
y
=
y =
y
=
f
(
x
)
f(x)
f
(
x
)
In the chain rule for implicit functions,
d
d
x
[
f
(
y
)
]
\frac{d}{dx} [f(y)]
d
x
d
[
f
(
y
)]
is equal to
f
′
(
y
)
⋅
d
y
d
x
f'(y) \cdot \frac{dy}{dx}
f
′
(
y
)
⋅
d
x
d
y
, where
d
y
d
x
\frac{dy}{dx}
d
x
d
y
represents the derivative
Implicit differentiation is used when both variables are dependent on
each
other.
Steps to perform implicit differentiation
1️⃣ Differentiate both sides of the equation
2️⃣ Use the chain rule for terms involving both variables
3️⃣ Solve for the derivative of the dependent variable
In an explicit function, one variable is expressed directly in terms of the
other
.
What is the general form of an implicit function?
F
(
x
,
y
)
=
F(x, y) =
F
(
x
,
y
)
=
0
0
0
In the example
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
, the derivative
d
y
d
x
\frac{dy}{dx}
d
x
d
y
is equal to
−
x
y
- \frac{x}{y}
−
y
x
Steps in implicit differentiation
1️⃣ Differentiate each term with respect to
x
x
x
2️⃣ Apply the chain rule where necessary
3️⃣ Solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
What is the purpose of checking the implicit differentiation solution with an explicit form?
To verify the derivative
What is
d
y
d
x
\frac{dy}{dx}
d
x
d
y
when differentiating
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
?
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
−
x
y
- \frac{x}{y}
−
y
x
Explicit differentiation is used when relationships are not easily expressed explicitly.
False
Steps for solving
d
y
d
x
\frac{dy}{dx}
d
x
d
y
in implicit differentiation
1️⃣ Isolate terms containing
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
2️⃣ Factor out
d
y
d
x
\frac{dy}{dx}
d
x
d
y
from those terms.
3️⃣ Divide both sides by the factor multiplying
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
Implicit differentiation is used when one variable is explicitly expressed as a function of the other.
False
Implicit differentiation is useful when the relationship between variables is not easily expressed
explicitly
Implicit functions require direct differentiation.
False
What is the first step in applying the chain rule to implicit functions?
Differentiate both sides
Solving for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
is the final step in applying the chain rule to implicit functions.
True
Explicit differentiation involves expressing one
variable
as a direct function of the other.
True
What is the form of an implicit function?
F
(
x
,
y
)
=
F(x, y) =
F
(
x
,
y
)
=
0
0
0
In an explicit function, one variable is expressed directly in terms of the
other
What is the chain rule formula for differentiating a composite function involving
y
y
y
?
d
d
x
[
f
(
y
)
]
=
\frac{d}{dx} [f(y)] =
d
x
d
[
f
(
y
)]
=
f
′
(
y
)
⋅
d
y
d
x
f'(y) \cdot \frac{dy}{dx}
f
′
(
y
)
⋅
d
x
d
y
The chain rule is used in implicit differentiation to account for terms involving
y
y
y
.
True
The final step in implicit differentiation is to isolate
d
y
d
x
\frac{dy}{dx}
d
x
d
y
When expressing
y
y
y
explicitly from the equation
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
, the positive root is \sqrt{25 - x^{2}}
Implicit and explicit differentiation methods produce the same derivative for the equation
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
.
True
Steps for implicit differentiation
1️⃣ Differentiate both sides of the equation with respect to
x
x
x
.
2️⃣ Apply the chain rule to terms involving
y
y
y
.
3️⃣ Solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
Implicit differentiation is essential for finding derivatives of
implicitly defined functions
.
True
d
d
x
(
y
2
)
=
\frac{d}{dx}(y^{2}) =
d
x
d
(
y
2
)
=
2
y
d
y
d
x
2y \frac{dy}{dx}
2
y
d
x
d
y
\frac{dy}{dx}
Isolating the term with \frac{dy}{dx}</latex> in
2
x
+
2x +
2
x
+
2
y
d
y
d
x
=
2y\frac{dy}{dx} =
2
y
d
x
d
y
=
0
0
0
results in
2
y
d
y
d
x
=
2y\frac{dy}{dx} =
2
y
d
x
d
y
=
−
2
x
- 2x
−
2
x
-2x
The explicit form of
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
is
y
=
y =
y
=
25
−
x
2
\sqrt{25 - x^{2}}
25
−
x
2
\sqrt{25}
Given the equation
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
25
25
25
, differentiating both sides with respect to
x
x
x
results in
2
x
+
2x +
2
x
+
2
y
d
y
d
x
=
2y\frac{dy}{dx} =
2
y
d
x
d
y
=
0
0
0
\frac{dy}{dx}
Solving for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
in
8
x
+
8x +
8
x
+
18
y
d
y
d
x
=
18y\frac{dy}{dx} =
18
y
d
x
d
y
=
0
0
0
gives
d
y
d
x
=
\frac{dy}{dx} =
d
x
d
y
=
−
4
x
9
y
- \frac{4x}{9y}
−
9
y
4
x
-
Implicit differentiation is the process of differentiating an equation with respect to a variable when both variables are
dependent
Steps to perform implicit differentiation
1️⃣ Differentiate both sides of the equation with respect to the independent variable.
2️⃣ Use the chain rule to differentiate terms where both variables are present.
3️⃣ Solve for
d
y
d
x
\frac{dy}{dx}
d
x
d
y
.
Match the function type with its example:
Explicit Function ↔️
y
=
y =
y
=
x
2
+
x^{2} +
x
2
+
3
x
+
3x +
3
x
+
1
1
1
Implicit Function ↔️
x
2
+
x^{2} +
x
2
+
y
2
=
y^{2} =
y
2
=
1
1
1
The chain rule is applied differently to implicit and
explicit
functions.
True
When using the chain rule for terms involving y</latex>, the formula is
d
d
x
[
f
(
y
)
]
=
\frac{d}{dx} [f(y)] =
d
x
d
[
f
(
y
)]
=
f
′
(
y
)
⋅
d
y
d
x
f'(y) \cdot \frac{dy}{dx}
f
′
(
y
)
⋅
d
x
d
y
, where
d
y
d
x
\frac{dy}{dx}
d
x
d
y
is the derivative of y
Match the feature with the type of function:
Relationship is direct ↔️ Explicit function
Relationship is indirect ↔️ Implicit function
What is the general form of an explicit function?
y
=
y =
y
=
f
(
x
)
f(x)
f
(
x
)
Implicit functions require implicit differentiation, while explicit functions use
direct differentiation
.
True
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