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AP Calculus AB
Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
3.3 Differentiating Inverse Functions
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An inverse function
f
−
1
(
x
)
f^{ - 1}(x)
f
−
1
(
x
)
of a function
f
(
x
)
f(x)
f
(
x
)
reverses the operation
What is a key property of inverse functions regarding their domain and range?
Domain of
f
f
f
is the range of
f
−
1
f^{ - 1}
f
−
1
An inverse function exists if f(x)</latex> is
one-to-one
True
The inverse function theorem provides the derivative of an inverse function at a point given the derivative of the original
function
The condition
f
′
(
a
)
≠
0
f'(a) \neq 0
f
′
(
a
)
=
0
is necessary for the inverse function theorem
True
If
f
(
a
)
=
f(a) =
f
(
a
)
=
b
b
b
, then
f
−
1
(
b
)
=
f^{ - 1}(b) =
f
−
1
(
b
)
=
a
a
a
True
The graph of
f
−
1
f^{ - 1}
f
−
1
is the reflection of
f
f
f
over the line
y
=
y =
y
=
x
x
x
True
How can you graphically determine if a function is one-to-one?
Horizontal line test
What is the formula for the derivative of an inverse function according to the inverse function theorem?
(
f
−
1
)
′
(
y
)
=
(f^{ - 1})'(y) =
(
f
−
1
)
′
(
y
)
=
1
f
′
(
a
)
\frac{1}{f'(a)}
f
′
(
a
)
1
What are the steps to find the inverse function
f
−
1
(
x
)
f^{ - 1}(x)
f
−
1
(
x
)
from a given function
f
(
x
)
f(x)
f
(
x
)
?
Replace
f
(
x
)
f(x)
f
(
x
)
with
y
y
y
, swap
x
x
x
and
y
y
y
, solve for
y
y
y
, replace
y
y
y
with
f
−
1
(
x
)
f^{ - 1}(x)
f
−
1
(
x
)
Inverse functions are functions that "undo" another
function
Match the property of inverse functions with its description:
Reflection ↔️ The graph of f^{-1} is the reflection of f over y = x
Identity ↔️ f(f^{-1}(x)) = x and f^{-1}(f(x)) = x
The inverse function theorem requires that
f
f
f
is differentiable and
f
′
(
x
)
≠
0
f'(x) \neq 0
f
′
(
x
)
=
0
.
True
Match the inverse trigonometric function with its derivative:
\arcsin x</latex> ↔️
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
arccos
x
\arccos x
arccos
x
↔️
−
1
1
−
x
2
\frac{ - 1}{\sqrt{1 - x^{2}}}
1
−
x
2
−
1
arctan
x
\arctan x
arctan
x
↔️
1
1
+
x
2
\frac{1}{1 + x^{2}}
1
+
x
2
1
Steps to find the derivative of
f
−
1
(
x
)
=
f^{ - 1}(x) =
f
−
1
(
x
)
=
arcsin
(
2
x
)
\arcsin(2x)
arcsin
(
2
x
)
1️⃣ Identify
f
(
x
)
=
f(x) =
f
(
x
)
=
sin
(
x
)
\sin(x)
sin
(
x
)
and
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
cos
(
x
)
\cos(x)
cos
(
x
)
2️⃣ Solve
sin
(
a
)
=
\sin(a) =
sin
(
a
)
=
2
x
2x
2
x
to find
a
=
a =
a
=
arcsin
(
2
x
)
\arcsin(2x)
arcsin
(
2
x
)
3️⃣ Calculate
f
′
(
arcsin
(
2
x
)
)
=
f'(\arcsin(2x)) =
f
′
(
arcsin
(
2
x
))
=
cos
(
arcsin
(
2
x
)
)
\cos(\arcsin(2x))
cos
(
arcsin
(
2
x
))
4️⃣ Use the inverse function theorem:
(
f
−
1
)
′
(
x
)
=
(f^{ - 1})'(x) =
(
f
−
1
)
′
(
x
)
=
1
cos
(
arcsin
(
2
x
)
)
\frac{1}{\cos(\arcsin(2x))}
c
o
s
(
a
r
c
s
i
n
(
2
x
))
1
Steps to apply the inverse function theorem
1️⃣ Find the derivative of the original function
2️⃣ Solve for
a
a
a
such that
f
(
a
)
=
f(a) =
f
(
a
)
=
x
x
x
3️⃣ Evaluate the derivative at
a
a
a
4️⃣ Apply the formula
(
f
−
1
)
′
(
x
)
=
(f^{ - 1})'(x) =
(
f
−
1
)
′
(
x
)
=
1
f
′
(
a
)
\frac{1}{f'(a)}
f
′
(
a
)
1
The domain of
f
f
f
is the range of
f
−
1
f^{ - 1}
f
−
1
, and vice versa.
True
What is the first step in applying the inverse function theorem to find the derivative of
f
−
1
(
y
)
f^{ - 1}(y)
f
−
1
(
y
)
?
Find
f
′
(
x
)
f'(x)
f
′
(
x
)
What is the formula for
(
f
−
1
)
′
(
y
)
(f^{ - 1})'(y)
(
f
−
1
)
′
(
y
)
in terms of
f
′
(
a
)
f'(a)
f
′
(
a
)
?
1
f
′
(
a
)
\frac{1}{f'(a)}
f
′
(
a
)
1
Match the inverse trigonometric function with its derivative:
arcsin
x
\arcsin x
arcsin
x
↔️
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
arccos
x
\arccos x
arccos
x
↔️
−
1
1
−
x
2
\frac{ - 1}{\sqrt{1 - x^{2}}}
1
−
x
2
−
1
arctan
x
\arctan x
arctan
x
↔️
1
1
+
x
2
\frac{1}{1 + x^{2}}
1
+
x
2
1
If f(x) = 2x + 3</latex>, what is
f
′
(
x
)
f'(x)
f
′
(
x
)
?
2
2
2
Steps to find the inverse function
f
−
1
(
x
)
f^{ - 1}(x)
f
−
1
(
x
)
from
f
(
x
)
f(x)
f
(
x
)
1️⃣ Replace
f
(
x
)
f(x)
f
(
x
)
with
y
y
y
2️⃣ Swap
x
x
x
and
y
y
y
3️⃣ Solve for
y
y
y
in terms of
x
x
x
4️⃣ Replace
y
y
y
with
f
−
1
(
x
)
f^{ - 1}(x)
f
−
1
(
x
)
f
(
f
−
1
(
x
)
)
=
f(f^{ - 1}(x)) =
f
(
f
−
1
(
x
))
=
x
x
x
and
f
−
1
(
f
(
x
)
)
=
f^{ - 1}(f(x)) =
f
−
1
(
f
(
x
))
=
x
x
x
are key relationships for inverse functions
Match the function with its inverse property:
Input
x
x
x
↔️ Output
y
y
y
Output
y
y
y
↔️ Input
x
x
x
The inverse function theorem requires that
f
f
f
is differentiable
True
The domain of
f
f
f
is the range of
f
−
1
f^{ - 1}
f
−
1
, and vice versa
What does an inverse function map the output of a function back to?
Original input
What condition must a function satisfy to have an inverse?
One-to-one
Steps to find (f^{ - 1})'(2)</latex> for
f
(
x
)
=
f(x) =
f
(
x
)
=
x
3
+
x^{3} +
x
3
+
1
1
1
1️⃣ Find
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
3
x
2
3x^{2}
3
x
2
2️⃣ Solve
f
(
a
)
=
f(a) =
f
(
a
)
=
2
2
2
to find
a
=
a =
a
=
1
1
1
3️⃣ Calculate
f
′
(
1
)
=
f'(1) =
f
′
(
1
)
=
3
3
3
4️⃣ Use the inverse function theorem:
(
f
−
1
)
′
(
2
)
=
(f^{ - 1})'(2) =
(
f
−
1
)
′
(
2
)
=
1
3
\frac{1}{3}
3
1
What is the derivative of
arcsin
(
2
x
)
\arcsin(2x)
arcsin
(
2
x
)
?
2
1
−
4
x
2
\frac{2}{\sqrt{1 - 4x^{2}}}
1
−
4
x
2
2
Applying the inverse function theorem:
(
f
−
1
)
′
(
x
)
=
(f^{ - 1})'(x) =
(
f
−
1
)
′
(
x
)
=
1
f
′
(
x
−
3
2
)
=
\frac{1}{f'(\frac{x - 3}{2})} =
f
′
(
2
x
−
3
)
1
=
1
2
\frac{1}{2}
2
1
2
What is an inverse function designed to do?
Reverse the operation
Match the function property with its description:
Domain ↔️ x values
Range ↔️ y values
Steps to find
(
f
−
1
)
′
(
2
)
(f^{ - 1})'(2)
(
f
−
1
)
′
(
2
)
for f(x) = x^{3} + 1</latex>
1️⃣ Find
f
′
(
x
)
=
f'(x) =
f
′
(
x
)
=
3
x
2
3x^{2}
3
x
2
2️⃣ Solve
f
(
a
)
=
f(a) =
f
(
a
)
=
2
2
2
, which gives
a
=
a =
a
=
1
1
1
3️⃣ Evaluate
f
′
(
1
)
=
f'(1) =
f
′
(
1
)
=
3
3
3
4️⃣ Apply the formula
(
f
−
1
)
′
(
2
)
=
(f^{ - 1})'(2) =
(
f
−
1
)
′
(
2
)
=
1
3
\frac{1}{3}
3
1
To find the value of
a
a
a
such that f(a) = y</latex>, you need to solve the equation f(a) =
y
y
y
What is the derivative of
arcsin
x
\arcsin x
arcsin
x
?
1
1
−
x
2
\frac{1}{\sqrt{1 - x^{2}}}
1
−
x
2
1
Steps to find the derivative of an inverse function using the inverse function theorem
1️⃣ Identify the original function
f
(
x
)
f(x)
f
(
x
)
and find its derivative
f
′
(
x
)
f'(x)
f
′
(
x
)
2️⃣ Find the value
x
=
x =
x
=
a
a
a
such that
f
(
a
)
=
f(a) =
f
(
a
)
=
y
y
y
3️⃣ Evaluate
f
′
(
a
)
f'(a)
f
′
(
a
)
4️⃣ Apply the formula
(
f
−
1
)
′
(
y
)
=
(f^{ - 1})'(y) =
(
f
−
1
)
′
(
y
)
=
1
f
′
(
a
)
\frac{1}{f'(a)}
f
′
(
a
)
1
A one-to-one function must pass the
horizontal line test
True
The domain of a function is the range of its inverse, and vice versa.
True
The inverse function theorem states that
(
f
−
1
)
′
(
y
)
=
(f^{ - 1})'(y) =
(
f
−
1
)
′
(
y
)
=
1
f
′
(
a
)
\frac{1}{f'(a)}
f
′
(
a
)
1
, where
f
′
(
a
)
≠
0
f'(a) \neq 0
f
′
(
a
)
=
0
and
y
=
y =
y
=
f
(
a
)
f(a)
f
(
a
)
.derivative
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