How do we obtain $df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ from $f(x,y)$?












0












$begingroup$


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:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?










share|cite|improve this question











$endgroup$








  • 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
















0












$begingroup$


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:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?










share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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:
$df = frac{partial f}{partial x} dx + frac{partial f}{partial y} dy$ ?










share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 18:25







user366312

















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














  • 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










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