Mixing
How does one interpret $dfrac{dx}{dy}$ for a function which isn't invertible?
$begingroup$
I was just going through the proof of derivative of inverse functions.
The statement reads:
If $y= f(x)$ is a differentiable function of $x$ such that it's inverse $x=f^{-1}(y)$ exists, then $x$ is a differentiable function of $y$ and it's derivative is $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}, dfrac{dy}{dx}≠0 $
Which naturally arises few questions, what if the inverse $y=f(x)$ doesn't exist?
My Questions :
- Does $dfrac{dx}{dy}$ still have a meaning?
- If so then what would it mean geometrically?
- Would $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}$ still hold? And if it does, then why do we even need the invertible condition in the statement?
calculus derivatives inverse-function
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
$endgroup$
add a comment |
$begingroup$
I was just going through the proof of derivative of inverse functions.
The statement reads:
If $y= f(x)$ is a differentiable function of $x$ such that it's inverse $x=f^{-1}(y)$ exists, then $x$ is a differentiable function of $y$ and it's derivative is $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}, dfrac{dy}{dx}≠0 $
Which naturally arises few questions, what if the inverse $y=f(x)$ doesn't exist?
My Questions :
- Does $dfrac{dx}{dy}$ still have a meaning?
- If so then what would it mean geometrically?
- Would $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}$ still hold? And if it does, then why do we even need the invertible condition in the statement?
calculus derivatives inverse-function
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
$endgroup$
$begingroup$
answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:24
$begingroup$
answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:26
$begingroup$
answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
$endgroup$
– Bijayan Ray
Jan 20 at 12:29
1
$begingroup$
$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
$endgroup$
– Sangchul Lee
Jan 20 at 12:41
$begingroup$
@WesleyStrik updated
$endgroup$
– William
Jan 20 at 13:08
add a comment |
$begingroup$
I was just going through the proof of derivative of inverse functions.
The statement reads:
If $y= f(x)$ is a differentiable function of $x$ such that it's inverse $x=f^{-1}(y)$ exists, then $x$ is a differentiable function of $y$ and it's derivative is $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}, dfrac{dy}{dx}≠0 $
Which naturally arises few questions, what if the inverse $y=f(x)$ doesn't exist?
My Questions :
- Does $dfrac{dx}{dy}$ still have a meaning?
- If so then what would it mean geometrically?
- Would $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}$ still hold? And if it does, then why do we even need the invertible condition in the statement?
calculus derivatives inverse-function
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
$endgroup$
I was just going through the proof of derivative of inverse functions.
The statement reads:
If $y= f(x)$ is a differentiable function of $x$ such that it's inverse $x=f^{-1}(y)$ exists, then $x$ is a differentiable function of $y$ and it's derivative is $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}, dfrac{dy}{dx}≠0 $
Which naturally arises few questions, what if the inverse $y=f(x)$ doesn't exist?
My Questions :
- Does $dfrac{dx}{dy}$ still have a meaning?
- If so then what would it mean geometrically?
- Would $dfrac{dx}{dy} = dfrac{1}{frac{dy}{dx}}$ still hold? And if it does, then why do we even need the invertible condition in the statement?
calculus derivatives inverse-function
calculus derivatives inverse-function
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
edited Jan 20 at 13:23
Martin Sleziak
44.8k10119272
44.8k10119272
asked Jan 20 at 12:13
William William
1,207414
asked Jan 20 at 12:13
William William
1,207414
asked Jan 20 at 12:13
William William
1,207414
1,207414
$begingroup$
answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:24
$begingroup$
answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:26
$begingroup$
answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
$endgroup$
– Bijayan Ray
Jan 20 at 12:29
1
$begingroup$
$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
$endgroup$
– Sangchul Lee
Jan 20 at 12:41
$begingroup$
@WesleyStrik updated
$endgroup$
– William
Jan 20 at 13:08
add a comment |
$begingroup$
answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:24
$begingroup$
answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:26
$begingroup$
answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
$endgroup$
– Bijayan Ray
Jan 20 at 12:29
1
$begingroup$
$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
$endgroup$
– Sangchul Lee
Jan 20 at 12:41
$begingroup$
@WesleyStrik updated
$endgroup$
– William
Jan 20 at 13:08
$begingroup$
answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:24
$begingroup$
answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:24
$begingroup$
answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:26
$begingroup$
answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
$endgroup$
– Bijayan Ray
Jan 20 at 12:26
$begingroup$
answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
$endgroup$
– Bijayan Ray
Jan 20 at 12:29
$begingroup$
answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
$endgroup$
– Bijayan Ray
Jan 20 at 12:29
1
1
$begingroup$
$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
$endgroup$
– Sangchul Lee
Jan 20 at 12:41
$begingroup$
$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
$endgroup$
– Sangchul Lee
Jan 20 at 12:41
$begingroup$
@WesleyStrik updated
$endgroup$
– William
Jan 20 at 13:08
$begingroup$
@WesleyStrik updated
$endgroup$
– William
Jan 20 at 13:08
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{frac13}$ exists, but is not differentiable at $y=0$.
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
$endgroup$
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
add a comment |
$begingroup$
Notationwise, you should interpret $$frac {dx}{dy}= frac{ df^{-1}(y)}{dy}$$
Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.
Take the parabola $y=x^2$, it does not have one unique inverse over $mathbb R$, it has two possible 'inverses' $pm sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
$endgroup$
$begingroup$
Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
$endgroup$
– William
Jan 20 at 13:24
$begingroup$
Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
|
show 1 more comment
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{frac13}$ exists, but is not differentiable at $y=0$.
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
$endgroup$
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
add a comment |
$begingroup$
Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{frac13}$ exists, but is not differentiable at $y=0$.
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
$endgroup$
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
add a comment |
$begingroup$
Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{frac13}$ exists, but is not differentiable at $y=0$.
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
$endgroup$
Your statement isn't actually true. For a counterexample, take $f(x)=x^3$. The inverse $f^{-1}(y)=y^{frac13}$ exists, but is not differentiable at $y=0$.
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
answered Jan 20 at 12:31
TonyKTonyK
42.9k356135
42.9k356135
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
add a comment |
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
$begingroup$
he updated, so $y=0$ is not allowed because then the derivative is zero ($x=0$).
$endgroup$
– Wesley Strik
Jan 20 at 14:29
add a comment |
$begingroup$
Notationwise, you should interpret $$frac {dx}{dy}= frac{ df^{-1}(y)}{dy}$$
Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.
Take the parabola $y=x^2$, it does not have one unique inverse over $mathbb R$, it has two possible 'inverses' $pm sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
$endgroup$
$begingroup$
Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
$endgroup$
– William
Jan 20 at 13:24
$begingroup$
Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
|
show 1 more comment
$begingroup$
Notationwise, you should interpret $$frac {dx}{dy}= frac{ df^{-1}(y)}{dy}$$
Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.
Take the parabola $y=x^2$, it does not have one unique inverse over $mathbb R$, it has two possible 'inverses' $pm sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
$endgroup$
$begingroup$
Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
$endgroup$
– William
Jan 20 at 13:24
$begingroup$
Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
|
show 1 more comment
$begingroup$
Notationwise, you should interpret $$frac {dx}{dy}= frac{ df^{-1}(y)}{dy}$$
Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.
Take the parabola $y=x^2$, it does not have one unique inverse over $mathbb R$, it has two possible 'inverses' $pm sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
$endgroup$
Notationwise, you should interpret $$frac {dx}{dy}= frac{ df^{-1}(y)}{dy}$$
Where we have defined $y = f(x)$ so $x = f^{-1}(y)$, now if the function $f$ is not invertible, we know that the inverse relation $f^{-1}$ is not a proper function even to begin with, so differentiation does not make an awful lot of sense.
Take the parabola $y=x^2$, it does not have one unique inverse over $mathbb R$, it has two possible 'inverses' $pm sqrt x$, this has two slopes for every $x$, opposite and equal and that's why we cannot define a unique derivative. We usually interpret $frac{dx}{dy}$ as the slope of the inverse function, but for non-invertible functions we run into problems. We simply do not understand what you mean by "the slope" - of what?
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
edited Jan 20 at 12:44
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
answered Jan 20 at 12:23
Wesley StrikWesley Strik
2,113423
2,113423
$begingroup$
Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
$endgroup$
– William
Jan 20 at 13:24
$begingroup$
Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
|
show 1 more comment
$begingroup$
Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
$endgroup$
– William
Jan 20 at 13:24
$begingroup$
Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
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– Wesley Strik
Jan 20 at 14:26
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I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
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– Wesley Strik
Jan 20 at 14:28
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If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
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– Wesley Strik
Jan 20 at 14:33
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Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
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– William
Jan 20 at 13:24
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Exactly but taking your example, $y= x^2$ if you differentiate wrt $y$, you'll have $dfrac{dx}{dy} = dfrac{1}{2x}$ and what does this mean?
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– William
Jan 20 at 13:24
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Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
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– Wesley Strik
Jan 20 at 14:21
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Geometrically you inverted the slope, so the tangent lines of $x^2$ have slope $2x$, whilste the inverse has the reciprocal, they always multiply together to give you 1, see: oregonstate.edu/instruct/mth251/cq/Stage6/Lesson/…
$endgroup$
– Wesley Strik
Jan 20 at 14:21
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I don't really think there is more to it, unless you start looking at advanced stuff like differentiable manifolds and stuff like that. From an analysis point of view it just tells you how to calculate the derivative of the inverse in terms of the derivative of the original function - which sometimes is a useful technique. Again, the inverse has a very clear meaning geometrically, you mirror the function in the line $y=x$ - and your question is about the slope of this, which to me is just the slope of the inverse, which is a function, so you are asking about the slope of a function.
$endgroup$
– Wesley Strik
Jan 20 at 14:26
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
I mean what made the original function so special? if a function is invertible, they are each other's inverses and they thus form a dual pair of functions with the property that composition with each other in any order gives the identity (group theory)
$endgroup$
– Wesley Strik
Jan 20 at 14:28
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
$begingroup$
If you do the substitution now in terms of $y$ we get that: $$ x= pm sqrt{y} implies frac{dx}{dy}= frac{1}{pm 2sqrt{y}}$$ which is precisely the slope of the 'inverse' functions $f(y)=pm sqrt{y}$
$endgroup$
– Wesley Strik
Jan 20 at 14:33
|
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answer 1 : do not exist but if piecewise inverse function exists then it exists for those partitions
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– Bijayan Ray
Jan 20 at 12:24
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answer 2: no meaning but for piecewise inverse function it has geometrical meaning as $frac {dx}{dy}$ within each partitions
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– Bijayan Ray
Jan 20 at 12:26
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answer 3: no $frac {dx}{dy} = frac {1}{frac {dy}{dx}}$ do not holds
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– Bijayan Ray
Jan 20 at 12:29
1
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$dx/dy$ can be interpreted as the coefficient of $dy$ for the differential form $dx$ over the submanifold ${(x,f(x)):xinmathbb{R}}$ of the Euclidean plane. More analytically, the graph of $y=f(x)$ may be invertible if we restrict its domain, and then one can still regard $dx/dy$ as the derivative of that 'local inverse'.
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– Sangchul Lee
Jan 20 at 12:41
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@WesleyStrik updated
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– William
Jan 20 at 13:08