How do we obtain $df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ from $f(x,y)$?
$begingroup$
See this: What is the Jacobian, how does it work, and what is an intuitive explanation of the Jacobian and a change of basis?
Consider a function $f(x,y)$. The differential $df$ is given by: $df =
frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$
This looks a lot like the gradient equation: $nabla f =
frac{partial f}{partial x} {i} + frac{partial f}{partial y}{ j}$
This is the basic idea behind the Jacobian (and differential forms) -
that a differential can be viewed as a sort of vector, with its
components being given by the partial derivative terms and the
‘basis’ as the differentials of the independent variables - $dx$ and $dy$.
... ...
How did he obtain:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?
differential
asked Dec 30 '18 at 8:01
user366312user366312
642317
$endgroup$
add a comment |
$begingroup$
See this: What is the Jacobian, how does it work, and what is an intuitive explanation of the Jacobian and a change of basis?
Consider a function $f(x,y)$. The differential $df$ is given by: $df =
frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$
This looks a lot like the gradient equation: $nabla f =
frac{partial f}{partial x} {i} + frac{partial f}{partial y}{ j}$
This is the basic idea behind the Jacobian (and differential forms) -
that a differential can be viewed as a sort of vector, with its
components being given by the partial derivative terms and the
‘basis’ as the differentials of the independent variables - $dx$ and $dy$.
... ...
How did he obtain:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?
differential
asked Dec 30 '18 at 8:01
user366312user366312
642317
$endgroup$
1
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
3
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43
add a comment |
$begingroup$
See this: What is the Jacobian, how does it work, and what is an intuitive explanation of the Jacobian and a change of basis?
Consider a function $f(x,y)$. The differential $df$ is given by: $df =
frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$
This looks a lot like the gradient equation: $nabla f =
frac{partial f}{partial x} {i} + frac{partial f}{partial y}{ j}$
This is the basic idea behind the Jacobian (and differential forms) -
that a differential can be viewed as a sort of vector, with its
components being given by the partial derivative terms and the
‘basis’ as the differentials of the independent variables - $dx$ and $dy$.
... ...
How did he obtain:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?
differential
asked Dec 30 '18 at 8:01
user366312user366312
642317
$endgroup$
See this: What is the Jacobian, how does it work, and what is an intuitive explanation of the Jacobian and a change of basis?
Consider a function $f(x,y)$. The differential $df$ is given by: $df =
frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$
This looks a lot like the gradient equation: $nabla f =
frac{partial f}{partial x} {i} + frac{partial f}{partial y}{ j}$
This is the basic idea behind the Jacobian (and differential forms) -
that a differential can be viewed as a sort of vector, with its
components being given by the partial derivative terms and the
‘basis’ as the differentials of the independent variables - $dx$ and $dy$.
... ...
How did he obtain:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?
differential
differential
asked Dec 30 '18 at 8:01
user366312user366312
642317
asked Dec 30 '18 at 8:01
user366312user366312
642317
edited Jan 4 at 18:25
user366312
asked Dec 30 '18 at 8:01
user366312user366312
642317
asked Dec 30 '18 at 8:01
user366312user366312
642317
asked Dec 30 '18 at 8:01
user366312user366312
642317
642317
1
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
3
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43
add a comment |
1
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
3
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43
1
1
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
3
3
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43
add a comment |
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1
$begingroup$
See The Chain Rule for Functions of Two Variables.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:09
3
$begingroup$
What do $df, dx$ and $dy$ mean to you? After considering that, maybe your problem is actually obvious? If not, then at least you have a much more concrete question to work on.
$endgroup$
– Arthur
Dec 30 '18 at 8:11
$begingroup$
The Chain rule is based on the principle of composition of functions. We have a function $f(x,y)$ and in turn $x$ and $y$ are function of a parameter $t$. What we have is : $f(x(t), y(t))$. Thus, in order to compute $dfrac {df}{dt}$ we have to apply the rule.
$endgroup$
– Mauro ALLEGRANZA
Dec 30 '18 at 8:20
$begingroup$
More or less the same thing was asked just a few hours ago: math.stackexchange.com/questions/3056517/…
$endgroup$
– Hans Lundmark
Dec 30 '18 at 8:43