Splitting partial derivatives











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How come
$$frac{partial f}{partial x} = frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x}$$
when
$$
u = x; cos theta - y; sin theta
$$
$$
v = x; sin theta + y; costheta
$$
It seems like that formula is twice as great on the right side. (Are the definitions of $u$ and $v$ even relevant?)



ANSWER: This is the definition of the chain rule for partial derivatives when $f$ is a function of $x$ and $y.$ It is irrelevant to the definitions of $u(x,y)$ and $v(x,y).$






calculus multivariable-calculus partial-derivative







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  • Use the chain rule for a real value function...
    – Curious Droid
    Oct 20 '14 at 8:57










  • Doesn't that play in to my question though about not being equal?
    – Ldaike Tren
    Oct 20 '14 at 8:59










  • You defined what $u$ and $v$ are, what about $f$?
    – 5xum
    Oct 20 '14 at 9:04










  • I was told the definition of $f$ is irrelevant.
    – Ldaike Tren
    Oct 20 '14 at 9:06










  • It is, but $f$ must be a funcion of $u$ and $v$.
    – 5xum
    Oct 20 '14 at 9:07















up vote
0
down vote

favorite












How come
$$frac{partial f}{partial x} = frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x}$$
when
$$
u = x; cos theta - y; sin theta
$$
$$
v = x; sin theta + y; costheta
$$
It seems like that formula is twice as great on the right side. (Are the definitions of $u$ and $v$ even relevant?)



ANSWER: This is the definition of the chain rule for partial derivatives when $f$ is a function of $x$ and $y.$ It is irrelevant to the definitions of $u(x,y)$ and $v(x,y).$






calculus multivariable-calculus partial-derivative







share|cite|improve this question
























  • Use the chain rule for a real value function...
    – Curious Droid
    Oct 20 '14 at 8:57










  • Doesn't that play in to my question though about not being equal?
    – Ldaike Tren
    Oct 20 '14 at 8:59










  • You defined what $u$ and $v$ are, what about $f$?
    – 5xum
    Oct 20 '14 at 9:04










  • I was told the definition of $f$ is irrelevant.
    – Ldaike Tren
    Oct 20 '14 at 9:06










  • It is, but $f$ must be a funcion of $u$ and $v$.
    – 5xum
    Oct 20 '14 at 9:07













up vote
0
down vote

favorite









up vote
0
down vote

favorite











How come
$$frac{partial f}{partial x} = frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x}$$
when
$$
u = x; cos theta - y; sin theta
$$
$$
v = x; sin theta + y; costheta
$$
It seems like that formula is twice as great on the right side. (Are the definitions of $u$ and $v$ even relevant?)



ANSWER: This is the definition of the chain rule for partial derivatives when $f$ is a function of $x$ and $y.$ It is irrelevant to the definitions of $u(x,y)$ and $v(x,y).$






calculus multivariable-calculus partial-derivative







share|cite|improve this question















How come
$$frac{partial f}{partial x} = frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x}$$
when
$$
u = x; cos theta - y; sin theta
$$
$$
v = x; sin theta + y; costheta
$$
It seems like that formula is twice as great on the right side. (Are the definitions of $u$ and $v$ even relevant?)



ANSWER: This is the definition of the chain rule for partial derivatives when $f$ is a function of $x$ and $y.$ It is irrelevant to the definitions of $u(x,y)$ and $v(x,y).$






calculus multivariable-calculus partial-derivative




calculus multivariable-calculus partial-derivative






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share|cite|improve this question













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edited Oct 20 '14 at 9:14

























asked Oct 20 '14 at 8:55









Ldaike Tren

85




85












  • Use the chain rule for a real value function...
    – Curious Droid
    Oct 20 '14 at 8:57










  • Doesn't that play in to my question though about not being equal?
    – Ldaike Tren
    Oct 20 '14 at 8:59










  • You defined what $u$ and $v$ are, what about $f$?
    – 5xum
    Oct 20 '14 at 9:04










  • I was told the definition of $f$ is irrelevant.
    – Ldaike Tren
    Oct 20 '14 at 9:06










  • It is, but $f$ must be a funcion of $u$ and $v$.
    – 5xum
    Oct 20 '14 at 9:07


















  • Use the chain rule for a real value function...
    – Curious Droid
    Oct 20 '14 at 8:57










  • Doesn't that play in to my question though about not being equal?
    – Ldaike Tren
    Oct 20 '14 at 8:59










  • You defined what $u$ and $v$ are, what about $f$?
    – 5xum
    Oct 20 '14 at 9:04










  • I was told the definition of $f$ is irrelevant.
    – Ldaike Tren
    Oct 20 '14 at 9:06










  • It is, but $f$ must be a funcion of $u$ and $v$.
    – 5xum
    Oct 20 '14 at 9:07
















Use the chain rule for a real value function...
– Curious Droid
Oct 20 '14 at 8:57




Use the chain rule for a real value function...
– Curious Droid
Oct 20 '14 at 8:57












Doesn't that play in to my question though about not being equal?
– Ldaike Tren
Oct 20 '14 at 8:59




Doesn't that play in to my question though about not being equal?
– Ldaike Tren
Oct 20 '14 at 8:59












You defined what $u$ and $v$ are, what about $f$?
– 5xum
Oct 20 '14 at 9:04




You defined what $u$ and $v$ are, what about $f$?
– 5xum
Oct 20 '14 at 9:04












I was told the definition of $f$ is irrelevant.
– Ldaike Tren
Oct 20 '14 at 9:06




I was told the definition of $f$ is irrelevant.
– Ldaike Tren
Oct 20 '14 at 9:06












It is, but $f$ must be a funcion of $u$ and $v$.
– 5xum
Oct 20 '14 at 9:07




It is, but $f$ must be a funcion of $u$ and $v$.
– 5xum
Oct 20 '14 at 9:07










1 Answer
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This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$



$$
df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
$$
I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So



$$
du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
$$



Substitute for these values of $du,dv$
$$
df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
$$



Group terms in dx, dy together and you have your result.



$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$






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    1 Answer
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    1 Answer
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    active

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    up vote
    0
    down vote













    This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$



    $$
    df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
    $$
    I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So



    $$
    du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
    dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
    $$



    Substitute for these values of $du,dv$
    $$
    df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
    $$



    Group terms in dx, dy together and you have your result.



    $$
    df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
    left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
    $$






    share|cite|improve this answer

























      up vote
      0
      down vote













      This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$



      $$
      df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
      $$
      I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So



      $$
      du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
      dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
      $$



      Substitute for these values of $du,dv$
      $$
      df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
      $$



      Group terms in dx, dy together and you have your result.



      $$
      df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
      left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
      $$






      share|cite|improve this answer























        up vote
        0
        down vote










        up vote
        0
        down vote









        This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$



        $$
        df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
        $$
        I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So



        $$
        du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
        dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
        $$



        Substitute for these values of $du,dv$
        $$
        df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
        $$



        Group terms in dx, dy together and you have your result.



        $$
        df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
        left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
        $$






        share|cite|improve this answer












        This is not a proof, but a demonstration: Had $f=f(u,v)$ only where u and v were independent variables, you'd get a total change in f from changes in both $u$ and $v$



        $$
        df = frac{partial f}{partial u} du + frac{partial f}{partial v} dv
        $$
        I suspect you are comfortable with that. As it turns out, $u$ and $v$ are themselves functions of independent variables x and y (assuming $theta$ is only a parameter here). So



        $$
        du = frac{partial u}{partial x} dx + frac{partial u}{partial y} dy \
        dv = frac{partial v}{partial x} dx + frac{partial v}{partial y} dy
        $$



        Substitute for these values of $du,dv$
        $$
        df = frac{partial f}{partial u} (frac{partial u}{partial x} dx + frac{partial u}{partial y} dy ) + frac{partial f}{partial v} (frac{partial v}{partial x} dx + frac{partial v}{partial y} dy )
        $$



        Group terms in dx, dy together and you have your result.



        $$
        df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
        left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
        $$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Oct 20 '14 at 9:24









        user_of_math

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