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If f(x)=e^x^4 find f^-1(x)and the derivative of f^-1(e^81).
Related Topics
Wize University Calculus 1 Textbook > Derivatives
Derivatives of Inverse Functions
3 Activities
If
f
(
x
)
=
e
x
4
f(x)=e^{x^4}
f
(
x
)
=
e
x
4
find
f
−
1
(
x
)
f^{-1}(x)
f
−
1
(
x
)
and the derivative of
f
−
1
(
e
81
)
f^{-1}(e^{81})
f
−
1
(
e
81
)
.
f
−
1
(
x
)
=
e
x
4
,
[
f
−
1
]
′
(
e
81
)
=
4
e
243
+
e
324
f^{-1}(x)=e^{x^4}, {[f^{-1}]}'(e^{81})=4e^{243+e^{324}}
f
−
1
(
x
)
=
e
x
4
,
[
f
−
1
]
′
(
e
81
)
=
4
e
243
+
e
324
f
−
1
(
x
)
=
(
ln
(
x
)
)
1
/
4
,
[
f
−
1
]
′
(
e
81
)
=
1
108
e
81
f^{-1}(x)=(\ln(x))^{1/4}, {[f^{-1}]}'(e^{81})=\frac{1}{108e^{81}}
f
−
1
(
x
)
=
(
ln
(
x
)
)
1/4
,
[
f
−
1
]
′
(
e
81
)
=
108
e
81
1
f
−
1
(
x
)
=
4
e
x
,
[
f
−
1
]
′
(
e
81
)
=
4
e
e
81
f^{-1}(x)=4 e^x, {[f^{-1}]}'(e^{81})=4e^{e^{81}}
f
−
1
(
x
)
=
4
e
x
,
[
f
−
1
]
′
(
e
81
)
=
4
e
e
81
f
−
1
(
x
)
=
(
ln
(
x
)
)
4
,
[
f
−
1
]
′
(
e
81
)
=
2125764
e
81
f^{-1}(x)=(\ln(x))^{4}, [f^{-1}]'(e^{81})=\frac{2125764}{e^{81}}
f
−
1
(
x
)
=
(
ln
(
x
)
)
4
,
[
f
−
1
]
′
(
e
81
)
=
e
81
2125764
I don't know
Check Submission
More Derivatives of Inverse Functions Questions:
Practice: Tangent to the Inverse
Q:
\textbf{Q:}
Q:
Let
f
(
x
)
=
x
3
f(x)=x^3
f
(
x
)
=
x
3
. Find the slope of the tangent line to
f
−
1
at
x
=
8
=
f
(
2
)
f^{-1}\text{ at }x=8=f(2)
f
−
1
at
x
=
8
=
f
(
2
)
.
Derivatives of Inverse Functions
Suppose
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
,
x
≥
3
f(x)=x^3-3x^2-x+4, x\geq 3
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
,
x
≥
3
. Assuming
f(x)
has an inverse
g(x)
, find the value of
g
′
(
16
)
g'(16)
g
′
(
16
)
, given that
f
(
4
)
=
16.
f(4)=16.
f
(
4
)
=
16.
Practice: Tangent to the Inverse
Q:
\textbf{Q:}
Q:
Let
f
(
x
)
=
x
3
f(x)=x^3
f
(
x
)
=
x
3
. Find the slope of the tangent line to
f
−
1
at
x
=
8
=
f
(
2
)
f^{-1}\text{ at }x=8=f(2)
f
−
1
at
x
=
8
=
f
(
2
)
.
Suppose
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
f(x)=x^3-3x^2-x+4
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
with
x
≥
3
x\geq 3
x
≥
3
. Assuming
f
(
x
)
f(x)
f
(
x
)
has an inverse
g
(
x
)
g(x)
g
(
x
)
, find the value of
g
′
(
16
)
g'(16)
g
′
(
16
)
, given that
f
(
4
)
=
16
f(4)=16
f
(
4
)
=
16
.
The function
f
(
x
)
=
x
3
+
x
+
1
f(x) = x^3 + x + 1
f
(
x
)
=
x
3
+
x
+
1
(where
x
>
0
x > 0
x
>
0
) has an inverse,
y
=
f
−
1
(
x
)
y=f^{-1}(x)
y
=
f
−
1
(
x
)
, on that interval. What is the slope of the curve
y
=
f
−
1
(
x
)
y = f^{-1}(x)
y
=
f
−
1
(
x
)
at the point
(
x
,
y
)
=
(
3
,
1
)
(x, y) = (3, 1)
(
x
,
y
)
=
(
3
,
1
)
?
Derivatives: Inverse Functions
Suppose
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
f(x) = x^3-3x^2-x+4
f
(
x
)
=
x
3
−
3
x
2
−
x
+
4
,
x
≥
3
x \geq 3
x
≥
3
. Assuming
f
(
x
)
f(x)
f
(
x
)
has an inverse
g
(
x
)
g(x)
g
(
x
)
, find the value of
g
′
(
16
)
g'(16)
g
′
(
16
)
, given that
f
(
4
)
=
16
f(4) = 16
f
(
4
)
=
16
.
Derivative of inverse functions
Let
f
(
x
)
=
x
3
+
4
x
−
2.
F
i
n
d
(
f
−
1
)
′
(
3
)
.
f(x)=x^3+4x−2.\ Find\ (f^{−1})'(3).
f
(
x
)
=
x
3
+
4
x
−
2.
F
in
d
(
f
−
1
)
′
(
3
)
.