Multivariate Chain Rule with a Function of one of the Variables
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I'm working through a particular problem that is giving me inconsistent results given the way I approach the problem, and I would like to confirm that I am appropriately understanding the chain rule in the case where my function, call it $z$, is a function of some $x(t)$ and some $y(x(t))$. That is, the function $z$ is a function of two variables, the second of which is a function of the first, or
$$z = f(x(t),y(x(t))).$$
I'm differentiating with respect to $t$ by applying the following rule:
$$frac{partial z}{partial t} = frac{partial z}{partial x}frac{partial x}{partial t} + frac{partial z}{partial y}frac{partial y}{partial x}frac{partial x}{partial t} = frac{partial x}{partial t}biggl[frac{partial z}{partial x} + frac{partial z}{partial y}frac{partial y}{partial x} biggr].$$
Is this the correct application of the chain rule in this case?
calculus derivatives chain-rule
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
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add a comment |
$begingroup$
I'm working through a particular problem that is giving me inconsistent results given the way I approach the problem, and I would like to confirm that I am appropriately understanding the chain rule in the case where my function, call it $z$, is a function of some $x(t)$ and some $y(x(t))$. That is, the function $z$ is a function of two variables, the second of which is a function of the first, or
$$z = f(x(t),y(x(t))).$$
I'm differentiating with respect to $t$ by applying the following rule:
$$frac{partial z}{partial t} = frac{partial z}{partial x}frac{partial x}{partial t} + frac{partial z}{partial y}frac{partial y}{partial x}frac{partial x}{partial t} = frac{partial x}{partial t}biggl[frac{partial z}{partial x} + frac{partial z}{partial y}frac{partial y}{partial x} biggr].$$
Is this the correct application of the chain rule in this case?
calculus derivatives chain-rule
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
$endgroup$
$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38
add a comment |
$begingroup$
I'm working through a particular problem that is giving me inconsistent results given the way I approach the problem, and I would like to confirm that I am appropriately understanding the chain rule in the case where my function, call it $z$, is a function of some $x(t)$ and some $y(x(t))$. That is, the function $z$ is a function of two variables, the second of which is a function of the first, or
$$z = f(x(t),y(x(t))).$$
I'm differentiating with respect to $t$ by applying the following rule:
$$frac{partial z}{partial t} = frac{partial z}{partial x}frac{partial x}{partial t} + frac{partial z}{partial y}frac{partial y}{partial x}frac{partial x}{partial t} = frac{partial x}{partial t}biggl[frac{partial z}{partial x} + frac{partial z}{partial y}frac{partial y}{partial x} biggr].$$
Is this the correct application of the chain rule in this case?
calculus derivatives chain-rule
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
$endgroup$
I'm working through a particular problem that is giving me inconsistent results given the way I approach the problem, and I would like to confirm that I am appropriately understanding the chain rule in the case where my function, call it $z$, is a function of some $x(t)$ and some $y(x(t))$. That is, the function $z$ is a function of two variables, the second of which is a function of the first, or
$$z = f(x(t),y(x(t))).$$
I'm differentiating with respect to $t$ by applying the following rule:
$$frac{partial z}{partial t} = frac{partial z}{partial x}frac{partial x}{partial t} + frac{partial z}{partial y}frac{partial y}{partial x}frac{partial x}{partial t} = frac{partial x}{partial t}biggl[frac{partial z}{partial x} + frac{partial z}{partial y}frac{partial y}{partial x} biggr].$$
Is this the correct application of the chain rule in this case?
calculus derivatives chain-rule
calculus derivatives chain-rule
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
asked Jan 10 at 3:22
ArchetuponArchetupon
13211
13211
$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38
add a comment |
$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38
$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38
add a comment |
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$begingroup$
This is correct.
$endgroup$
– NicNic8
Jan 10 at 4:15
$begingroup$
It look right to me.
$endgroup$
– callculus
Jan 10 at 4:38