Partial Derivatives: Chain Rule Confirmation

[fishingtreeing]

hello, i am supposed to use the two variable chain rule to confirm that changing variables from (x,y) to (v,w) with v=x and w=y/x leads to:

\partial{v}=\partial{x} + w\partial{y}
and \partial{w}=x\partial{y}

it seems to me that the first line should read \partial{v}=\partial{x} = w\partial{y}

is this true? i am a bit rusty on multivariable calculus

 

[dextercioby]

Please,use the [ tex ] and [ /tex ] tags (without the spaces,of course).It doesn't make too much sense what you've written there.

Usually $\partial D$ all by himself means the boundary of the domain D.

Daniel.

 

[quetzalcoatl9]

hello, i am supposed to use the two variable chain rule to confirm that changing variables from (x,y) to (v,w) with v=x and w=y/x leads to:

\partial{v}=\partial{x} + w\partial{y}
and \partial{w}=x\partial{y}

it seems to me that the first line should read \partial{v}=\partial{x} = w\partial{y}

is this true? i am a bit rusty on multivariable calculus

for this to make any sense, you need to express it as a function, say $f(x,y)$ and then take derivatives of $f$

 

[fishingtreeing]

ok, here is a repost

hello, i am supposed to use the two variable chain rule to confirm that changing variables from (x,y) to (v,w) with v=x and w=y/x leads to:

$\frac{\partial}{\partial v} = \frac{\partial}{\partial x} + w\frac{\partial}{\partial y}$
and $\frac{\partial}{\partial w}=x\frac{\partial}{\partial y}$

it seems to me that the first line should read $\frac{\partial}{\partial v} = \frac{\partial}{\partial x} = w\frac{\partial}{\partial y}$

is this true? i am a bit rusty on multivariable calculus

 

[dextercioby]

Hmmm

$$\frac{\partial}{\partial v}=\frac{\partial x}{\partial v}\frac{\partial}{\partial x}+ \frac{\partial y}{\partial v}\frac{\partial}{\partial y}$$

$$\frac{\partial}{\partial w}=\frac{\partial x}{\partial w}\frac{\partial}{\partial x}+\frac{\partial y}{\partial w}\frac{\partial}{\partial y}$$

There's no other way to apply the chain rule.

You could put it as well in matrix notation using the Jacobian matrix.

Daniel.

 

[fishingtreeing]

ok, thanks
i can see where the first one comes from, but if i apply the chain rule to the second one, i get

$\frac{\partial}{\partial w}=\frac{-y}{w^2}\frac{\partial}{\partial x} + x\frac{\partial}{\partial y}$

is this correct, or is the original correct?