Derivative of an inverse of a function [closed]
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I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.
calculus functions derivatives
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
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closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.
calculus functions derivatives
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
$endgroup$
closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
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– Martin R
Jan 20 at 13:39
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Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
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– Eric Lippert
Jan 20 at 17:37
add a comment |
$begingroup$
I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.
calculus functions derivatives
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
$endgroup$
I'm new into Calculus. Given $f:mathbb{R}rightarrow mathbb{R}$, I know that $f(0)=2$ and $f'(0)=3$. I can't understand why, given a function $g=f^{-1}$, it is true that $g'(2)=frac{1}{3}$.
calculus functions derivatives
calculus functions derivatives
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
asked Jan 20 at 13:35
Kevin MooreKevin Moore
16918
16918
closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory Jan 20 at 19:19
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – TheSimpliFire, Did, José Carlos Santos, RRL, steven gregory
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39
$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37
add a comment |
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39
$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39
$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37
$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37
add a comment |
3 Answers
3
active
oldest
votes
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The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$
You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:
It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
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add a comment |
$begingroup$
Explanation without words [oops, was that too many words?]. 
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
$endgroup$
add a comment |
$begingroup$
Let's try :
$g=f^{-1}$;
$g(2)=f^{-1}(f(0))=0.$
$g(f(x))= x;$
$g'(f(x))f'(x)=1;$
(Chain rule)
$g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.
$g'(f(0))= frac{1}{f'(0)};$
$g'(2)=frac{1}{3};$
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$
You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:
It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
$endgroup$
add a comment |
$begingroup$
The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$
You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:
It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
$endgroup$
add a comment |
$begingroup$
The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$
You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:
It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
$endgroup$
The idea is to write
$$
f(g(x))=x
$$
then use the chain rule
$$
f'(g(x))g'(x)=1
$$
From this you get:
$$
g'(x)=frac{1}{f'(g(x))}
$$
With your numerical example: $f'(0)=3$ and $f(0)=2Rightarrow g(2)=0$, therefore $g'(2)=frac{1}{3}$
You can have a look at Inverse_functions_and_differentiation (wiki page) to get a more detailed discussion. One important and useful property to remember is:
It follows that a function that has a continuous derivative has an
inverse in a neighbourhood of every point where the derivative is
non-zero. This need not be true if the derivative is not continuous
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
edited Jan 20 at 13:55
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
answered Jan 20 at 13:47
Picaud VincentPicaud Vincent
1,52439
1,52439
add a comment |
add a comment |
$begingroup$
Explanation without words [oops, was that too many words?].
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
$endgroup$
add a comment |
$begingroup$
Explanation without words [oops, was that too many words?].
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
$endgroup$
add a comment |
$begingroup$
Explanation without words [oops, was that too many words?].
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
$endgroup$
Explanation without words [oops, was that too many words?].
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
answered Jan 20 at 14:23
John JoyJohn Joy
6,32011727
6,32011727
add a comment |
add a comment |
$begingroup$
Let's try :
$g=f^{-1}$;
$g(2)=f^{-1}(f(0))=0.$
$g(f(x))= x;$
$g'(f(x))f'(x)=1;$
(Chain rule)
$g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.
$g'(f(0))= frac{1}{f'(0)};$
$g'(2)=frac{1}{3};$
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Let's try :
$g=f^{-1}$;
$g(2)=f^{-1}(f(0))=0.$
$g(f(x))= x;$
$g'(f(x))f'(x)=1;$
(Chain rule)
$g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.
$g'(f(0))= frac{1}{f'(0)};$
$g'(2)=frac{1}{3};$
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
add a comment |
$begingroup$
Let's try :
$g=f^{-1}$;
$g(2)=f^{-1}(f(0))=0.$
$g(f(x))= x;$
$g'(f(x))f'(x)=1;$
(Chain rule)
$g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.
$g'(f(0))= frac{1}{f'(0)};$
$g'(2)=frac{1}{3};$
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
$endgroup$
Let's try :
$g=f^{-1}$;
$g(2)=f^{-1}(f(0))=0.$
$g(f(x))= x;$
$g'(f(x))f'(x)=1;$
(Chain rule)
$g'(f(x))=frac{1}{f'(x)},$ $(f'(x) not =0)$.
$g'(f(0))= frac{1}{f'(0)};$
$g'(2)=frac{1}{3};$
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
edited Jan 20 at 16:13
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
answered Jan 20 at 13:44
Peter SzilasPeter Szilas
11.3k2822
11.3k2822
add a comment |
add a comment |
$begingroup$
Did you have a look at en.wikipedia.org/wiki/Inverse_functions_and_differentiation ?
$endgroup$
– Martin R
Jan 20 at 13:39
$begingroup$
Take a look at some functions that have inverses; draw both the functions and their inverses on the same graph. Do you notice anything symmetrical about the graph you just drew?
$endgroup$
– Eric Lippert
Jan 20 at 17:37