Multi-variable chain rule with multi-variable functions as arguments

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What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?



Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.



Is the general form of multi-variable chain rule similar to the following?



$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$



Thanks.







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  • I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
    – R zu
    Aug 28 at 16:23











  • Thanks for the answers. You have my eternal gratitude.
    – R zu
    Aug 28 at 17:04














up vote
1
down vote

favorite












What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?



Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.



Is the general form of multi-variable chain rule similar to the following?



$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$



Thanks.







share|cite|improve this question






















  • I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
    – R zu
    Aug 28 at 16:23











  • Thanks for the answers. You have my eternal gratitude.
    – R zu
    Aug 28 at 17:04












up vote
1
down vote

favorite









up vote
1
down vote

favorite











What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?



Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.



Is the general form of multi-variable chain rule similar to the following?



$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$



Thanks.







share|cite|improve this question














What is the chain rule of a multi-variable function with arguments that are also multi-variable functions?



Suppose $x$, $y$, $z$ are independent variables. I mean changing $x$ won't change $y$ and $z$.



Is the general form of multi-variable chain rule similar to the following?



$$fracpartial w(u(x, y), v(y, z), t(y, z))partial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x $$



Thanks.









share|cite|improve this question













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edited Aug 28 at 16:25









StubbornAtom

4,08511135




4,08511135










asked Aug 28 at 16:21









R zu

1587




1587











  • I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
    – R zu
    Aug 28 at 16:23











  • Thanks for the answers. You have my eternal gratitude.
    – R zu
    Aug 28 at 17:04
















  • I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
    – R zu
    Aug 28 at 16:23











  • Thanks for the answers. You have my eternal gratitude.
    – R zu
    Aug 28 at 17:04















I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
– R zu
Aug 28 at 16:23





I am sure a gazillion people already know the answer... but all I find on the web is the $dx/dt$ thing and all have single variable function as argument of another function. Please tell me if you know this. Thanks.
– R zu
Aug 28 at 16:23













Thanks for the answers. You have my eternal gratitude.
– R zu
Aug 28 at 17:04




Thanks for the answers. You have my eternal gratitude.
– R zu
Aug 28 at 17:04










2 Answers
2






active

oldest

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up vote
3
down vote



accepted










That's nearly right, but you left off the third term that accounts for $t$. So you should have
$$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$



where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.



One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.




Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.






share|cite|improve this answer






















  • I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
    – R zu
    Aug 28 at 17:05

















up vote
1
down vote













beginalign
fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x,
endalign
or what I think you meant to write
beginalign
fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
&=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
endalign






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    3
    down vote



    accepted










    That's nearly right, but you left off the third term that accounts for $t$. So you should have
    $$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$



    where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.



    One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.




    Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.






    share|cite|improve this answer






















    • I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
      – R zu
      Aug 28 at 17:05














    up vote
    3
    down vote



    accepted










    That's nearly right, but you left off the third term that accounts for $t$. So you should have
    $$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$



    where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.



    One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.




    Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.






    share|cite|improve this answer






















    • I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
      – R zu
      Aug 28 at 17:05












    up vote
    3
    down vote



    accepted







    up vote
    3
    down vote



    accepted






    That's nearly right, but you left off the third term that accounts for $t$. So you should have
    $$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$



    where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.



    One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.




    Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.






    share|cite|improve this answer














    That's nearly right, but you left off the third term that accounts for $t$. So you should have
    $$fracpartial wpartial x = fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x + fracpartial wpartial tcdotfracpartial tpartial x$$



    where $w$ is a function of $u,v,t$ and $u,v,t$ are functions of $x,y,z$.



    One mnemonic device for remembering this is to think of it as summing over all products of "fractions" that will partially cancel and each give $partial w$ in the numerator and $partial x$ in the denominator. But be warned that that's not really what's happening, these aren't fractions and they aren't cancelling. Nonetheless, this helps you to remember it.




    Addendum: Of course, if $x$ doesn't actually appear in the formula for $u$, $v$, or $t$, then that partial is zero. What I wrote is the most general formula, not just the formula for your specific case. In other words, even if $v=v(y,z)$ , you can still think of it as $v(x,y,z)$; but then $partial v/partial x = 0$ , and similarly for the others. Not sure if you really wrote what you meant, but my answer is applicable to the general case.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Aug 28 at 17:19

























    answered Aug 28 at 16:46









    MPW

    28.6k11853




    28.6k11853











    • I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
      – R zu
      Aug 28 at 17:05
















    • I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
      – R zu
      Aug 28 at 17:05















    I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
    – R zu
    Aug 28 at 17:05




    I saw a linear algebra proof of the simple chain rule: proofwiki.org/wiki/Chain_Rule_for_Real-Valued_Functions But I haven't studied that yet. Thanks a lot for the answer.
    – R zu
    Aug 28 at 17:05










    up vote
    1
    down vote













    beginalign
    fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
    &=fracpartial wpartial ucdotfracpartial upartial x,
    endalign
    or what I think you meant to write
    beginalign
    fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
    &=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
    endalign






    share|cite|improve this answer
























      up vote
      1
      down vote













      beginalign
      fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
      &=fracpartial wpartial ucdotfracpartial upartial x,
      endalign
      or what I think you meant to write
      beginalign
      fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
      &=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
      endalign






      share|cite|improve this answer






















        up vote
        1
        down vote










        up vote
        1
        down vote









        beginalign
        fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
        &=fracpartial wpartial ucdotfracpartial upartial x,
        endalign
        or what I think you meant to write
        beginalign
        fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
        &=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
        endalign






        share|cite|improve this answer












        beginalign
        fracpartial w(u(x, y), v(y, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotoverbracefracpartial vpartial x^0+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
        &=fracpartial wpartial ucdotfracpartial upartial x,
        endalign
        or what I think you meant to write
        beginalign
        fracpartial w(u(x, y), v(colorredx, z), t(y, z))partial x &= fracpartial wpartial ucdotfracpartial upartial x + fracpartial wpartial vcdotfracpartial vpartial x+ fracpartial wpartial tcdotoverbracefracpartial tpartial x^0 \
        &=fracpartial wpartial ucdotfracpartial upartial x+ fracpartial wpartial vcdotfracpartial vpartial x.
        endalign







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 28 at 16:53









        mzp

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