Splitting partial derivatives
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0
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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
asked Oct 20 '14 at 8:55
Ldaike Tren
85
|
show 2 more comments
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
asked Oct 20 '14 at 8:55
Ldaike Tren
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
|
show 2 more comments
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
asked Oct 20 '14 at 8:55
Ldaike Tren
85
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
asked Oct 20 '14 at 8:55
Ldaike Tren
85
asked Oct 20 '14 at 8:55
Ldaike Tren
85
edited Oct 20 '14 at 9:14
asked Oct 20 '14 at 8:55
Ldaike Tren
85
asked Oct 20 '14 at 8:55
Ldaike Tren
85
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
|
show 2 more comments
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
|
show 2 more comments
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.
$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$
answered Oct 20 '14 at 9:24
user_of_math
3,263727
add a comment |
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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.
$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$
answered Oct 20 '14 at 9:24
user_of_math
3,263727
add a comment |
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.
$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$
answered Oct 20 '14 at 9:24
user_of_math
3,263727
add a comment |
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.
$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$
answered Oct 20 '14 at 9:24
user_of_math
3,263727
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.
$$
df = left(frac{partial f}{partial u} frac{partial u}{partial x} + frac{partial f}{partial v} frac{partial v}{partial x} right) dx +
left(frac{partial f}{partial u} frac{partial u}{partial y} + frac{partial f}{partial v} frac{partial v}{partial y} right) dy
$$
answered Oct 20 '14 at 9:24
user_of_math
3,263727
answered Oct 20 '14 at 9:24
user_of_math
3,263727
answered Oct 20 '14 at 9:24
user_of_math
3,263727
answered Oct 20 '14 at 9:24
user_of_math
3,263727
3,263727
add a comment |
add a comment |
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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