Use chain rule to find expression for $frac{partial f}{partial t}$
$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$
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
asked Nov 25 at 0:57
AColoredReptile
1788
add a comment |
$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$
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
asked Nov 25 at 0:57
AColoredReptile
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
add a comment |
$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$
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
asked Nov 25 at 0:57
AColoredReptile
1788
$f(x,y,z,t)$, $x(t)$, $y(x,t,s)$, $z(y,x)$
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
asked Nov 25 at 0:57
AColoredReptile
1788
asked Nov 25 at 0:57
AColoredReptile
1788
edited Nov 26 at 0:45
asked Nov 25 at 0:57
AColoredReptile
1788
asked Nov 25 at 0:57
AColoredReptile
1788
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
add a comment |
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
add a comment |
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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