Use chain rule to find expression for $frac{partial f}{partial t}$












1














$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$



So differentiating $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}frac{partial y}{partial t}+frac{partial f}{partial z}frac{partial z}{partial t}+frac{partial f}{partial t}$



then differentiating y and z with respect to $t$, $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}]+frac{partial f}{partial z}[frac{partial z}{partial x}frac{partial x}{partial t}{frac{partial z}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}}]]+frac{partial f}{partial t}$



Is this correct?






multivariable-calculus







share|cite|improve this question
























  • Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
    – AColoredReptile
    Nov 25 at 1:26










  • For the partial wrt z there are more terms as y and x bear dependence on t.
    – Rafa Budría
    Nov 25 at 10:40










  • @Rafa Budría Ah I fixed it.
    – AColoredReptile
    Nov 26 at 0:45










  • Still a plus sign into the derivative for $z$ and you are done :)
    – Rafa Budría
    Nov 26 at 7:19
















1














$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$



So differentiating $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}frac{partial y}{partial t}+frac{partial f}{partial z}frac{partial z}{partial t}+frac{partial f}{partial t}$



then differentiating y and z with respect to $t$, $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}]+frac{partial f}{partial z}[frac{partial z}{partial x}frac{partial x}{partial t}{frac{partial z}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}}]]+frac{partial f}{partial t}$



Is this correct?






multivariable-calculus







share|cite|improve this question
























  • Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
    – AColoredReptile
    Nov 25 at 1:26










  • For the partial wrt z there are more terms as y and x bear dependence on t.
    – Rafa Budría
    Nov 25 at 10:40










  • @Rafa Budría Ah I fixed it.
    – AColoredReptile
    Nov 26 at 0:45










  • Still a plus sign into the derivative for $z$ and you are done :)
    – Rafa Budría
    Nov 26 at 7:19














1












1








1







$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$



So differentiating $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}frac{partial y}{partial t}+frac{partial f}{partial z}frac{partial z}{partial t}+frac{partial f}{partial t}$



then differentiating y and z with respect to $t$, $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}]+frac{partial f}{partial z}[frac{partial z}{partial x}frac{partial x}{partial t}{frac{partial z}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}}]]+frac{partial f}{partial t}$



Is this correct?






multivariable-calculus







share|cite|improve this question















$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$



So differentiating $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}frac{partial y}{partial t}+frac{partial f}{partial z}frac{partial z}{partial t}+frac{partial f}{partial t}$



then differentiating y and z with respect to $t$, $frac{partial f}{partial t}= frac{partial f}{partial x}frac{partial x}{partial t}+frac{partial f}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}]+frac{partial f}{partial z}[frac{partial z}{partial x}frac{partial x}{partial t}{frac{partial z}{partial y}[frac{partial y}{partial x}frac{partial x}{partial t}+frac{partial y}{partial t}}]]+frac{partial f}{partial t}$



Is this correct?






multivariable-calculus




multivariable-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 26 at 0:45

























asked Nov 25 at 0:57









AColoredReptile

1788




1788












  • Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
    – AColoredReptile
    Nov 25 at 1:26










  • For the partial wrt z there are more terms as y and x bear dependence on t.
    – Rafa Budría
    Nov 25 at 10:40










  • @Rafa Budría Ah I fixed it.
    – AColoredReptile
    Nov 26 at 0:45










  • Still a plus sign into the derivative for $z$ and you are done :)
    – Rafa Budría
    Nov 26 at 7:19


















  • Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
    – AColoredReptile
    Nov 25 at 1:26










  • For the partial wrt z there are more terms as y and x bear dependence on t.
    – Rafa Budría
    Nov 25 at 10:40










  • @Rafa Budría Ah I fixed it.
    – AColoredReptile
    Nov 26 at 0:45










  • Still a plus sign into the derivative for $z$ and you are done :)
    – Rafa Budría
    Nov 26 at 7:19
















Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
– AColoredReptile
Nov 25 at 1:26




Yeah I wasn't sure. Usually when I've seen questions like this it's something like $f(x(t),y(t))$ so I wasn't sure what to do when $t$ is a component of the function.
– AColoredReptile
Nov 25 at 1:26












For the partial wrt z there are more terms as y and x bear dependence on t.
– Rafa Budría
Nov 25 at 10:40




For the partial wrt z there are more terms as y and x bear dependence on t.
– Rafa Budría
Nov 25 at 10:40












@Rafa Budría Ah I fixed it.
– AColoredReptile
Nov 26 at 0:45




@Rafa Budría Ah I fixed it.
– AColoredReptile
Nov 26 at 0:45












Still a plus sign into the derivative for $z$ and you are done :)
– Rafa Budría
Nov 26 at 7:19




Still a plus sign into the derivative for $z$ and you are done :)
– Rafa Budría
Nov 26 at 7:19















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