Implicit partial differentiation

[chetzread]

Homework Statement

in the notes , 'by applying chain rule to LHS of the above equation ' , which equation is the author referring to ?
it's given that
f /x + (f/z)(z/x) = 0 ,
As we can see , the function contain variable x , y and z

Homework Equations

The Attempt at a Solution

why not
f /x + (f/z)(z/x) +(f/y)(y/x) = 0 ?

 

[BvU]

$f(x,y,z) = 0$

 

[BvU]

why not
$$f /x + (f/z)(z/x) +(f/y)(y/x) = 0 \ \ \ ?$$

because that is not true. That is dividing, not differentiating.

If you mean "why not$$
{df\over dx} = f_x + f_y y_x + f_z z_x$$ then the answer is: that's exactly what the author is doing.

 

[BvU]

Given up on the other thread ?

 

[BvU]

because that is not true. That is dividing, not differentiating.

If you mean "why not$$
{df\over dx} = f_x + f_y y_x + f_z z_x$$ then the answer is: that's exactly what the author is doing.

Note: The author makes a notation mistake (he/she too !). It should have read:

Differentiate (*) both sides wrt $x$, $$
{df\over dx} = {d\over dx} (0) $$​

Do you see the difference ? Do you understand what is meant with a partial derivative $\partial f\over\partial x$ (often written as $f_x$) as opposed to a total derivative $df\over dx$ ?

 

[BvU]

it's given that
f /x + (f/z)(z/x) = 0 ,

First:

not $\quad f /x + (f/z)(z/x) = 0\quad$ but $\quad f _x + f_z \;\displaystyle {\partial z\over\partial x} = 0$​

Second:

It's not given, but derived from $z=f(x,y)$​

 

[chetzread]

Given up on the other thread ?

no, i have attached the working in the previous thread, still waiting for your reply

 

[chetzread]

because that is not true. That is dividing, not differentiating.

If you mean "why not$$
{df\over dx} = f_x + f_y y_x + f_z z_x$$ then the answer is: that's exactly what the author is doing.

well , i couldn't understand why the author wrote ∂f/∂x + (∂f/∂z)(∂z/∂x) = 0 , i have no idea how he come up with this equation...In the notes, he said that byapplying chain rule to LHS of equation? which equation did he mean?

 

[Ray Vickson]

Homework Statement

in the notes , 'by applying chain rule to LHS of the above equation ' , which equation is the author referring to ?
it's given that
f /x + (f/z)(z/x) = 0 ,
As we can see , the function contain variable x , y and z

Homework Equations

The Attempt at a Solution

why not
f /x + (f/z)(z/x) +(f/y)(y/x) = 0 ?

You may be confused partly because the cited article writes confusingly. You can't have both $f(x,y,z) = 0$ and $z = f(x,y)$: it is absolutely never allowed to use the same symbol $``f"$ to stand for two totally different functions in the same sentence or two. I am surprised the author would make such a blunder.

Oh well, maybe that's just me being grumpy this morning.

 

[chetzread]

because that is not true. That is dividing, not differentiating.

If you mean "why not$$
{df\over dx} = df_dx +d f_dy (dy_dx) + (df_dz)(d z_dx) $$ then the answer is: that's exactly what the author is doing.

do you mean the equation should look like this ?
$${df\over dx} = df/dx +d f/dy (dy/dx) + (df/dz)(d z/dx) $$

why the LHS is
( df /dx ) shouldn't it = 0 ? , just like the author's working ?

 

[BvU]

well , i couldn't understand why the author wrote ∂f/∂x + (∂f/∂z)(∂z/∂x) = 0 , i have no idea how he come up with this equation...In the notes, he said that by applying chain rule to LHS of equation? which equation did he mean?

We'll humour mfb and write$$f(x,y,z) = f\left (x,y,z\left (x,y\right )\right ) = 0$$ Apply the definition of derivative $$ {dg\over dx} \equiv\ \ \lim\limits_{h\downarrow 0} {g(x+h)-g(x)\over h}$$while y, as the other independent variable, is kept constant. Since $f=0$ and stays 0, this derivative should give 0.

this is how the author was lured into writing $\displaystyle {\partial f\over\partial x}$
and I don't know how to do it didactically foolproof either. The point is that y is kept constant (because it is an independent variable) but z is a function of x so z does vary -- and thereby influences the derivative of f wrt x.​

So what we get is $$ 0 =
\ \ \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x,y))\over h} = \\
\quad \lim\limits_{h\downarrow 0} {f(x+h, y, z(x+h,y)) - f(x,y,z(x+h,y) )\over h} + {f(x, y, z(x+h,y)) - f(x,y,z(x,y) )\over h}
$$
First term is $\displaystyle {\partial f\over\partial x}$ a.k.a. $f_x$

Second term is $\displaystyle {\partial f\over\partial z} \displaystyle {\partial z\over\partial x}$ a.k.a. $f_z z_x$
In summary: $$f_x + f_z z_x = 0$$

Look up chain rule in your textbook.

 

[BvU]

do you mean the equation should look like this ?
$${df\over dx} = df/dx +d f/dy (dy/dx) + (df/dz)(d z/dx) $$

why the LHS is
( df /dx ) shouldn't it = 0 ? , just like the author's working ?

quoting a quote looks messy in PF. Can you please distinguish between $d$ and $\partial$, otherwise we might as well type chinese characters.

The author confuses you ( and irritates mfb and me also) by using the same $\partial f\over \partial x$ for two different things:

If he/she writes ${\partial f\over \partial x} = {\partial \over \partial x} (0)$ (which is 0)

that means: while keeping y the same and letting z vary with x
(it is a partial derivative because f is a function of two independent variables).​

If he/she writes ${\partial f\over \partial x} + {\partial f\over \partial z} {\partial z\over \partial x } =0$

that means: while keeping y and z the same. Now f is considered as a function of three ' independent' variables.
​

Drives one crazy, I agree. Still, it's elementary.

 

[BvU]

no, i have attached the working in the previous thread, still waiting for your reply

Oh, missed the alert. Will look.

Nope, you missed the alert: I posted #67 with more 'suggestions for improvement'

 

[chetzread]

can you explain the working? why we need to break the working into 2 parts?

 

[chetzread]

If he/she writes ∂f∂x+∂f∂z∂z∂x=0∂f∂x+∂f∂z∂z∂x=0 {\partial f\over \partial x} + {\partial f\over \partial z} {\partial z\over \partial x } =0
that means: while keeping y and z the same. Now f is considered as a function of three ' independent' variables.

well you said that we keep y and z same now...Why there's (∂f/∂z)(∂z/∂x)appear in equation...
(∂f/∂z)(∂z/∂x) mean function f(x,y,z) varies with z,am i right?

 

[BvU]

can you explain the working? why we need to break the working into 2 parts?

To show you how the two terms of $\displaystyle {\partial f\over \partial x}$, namely $\ f_x\$ and $\ f_z \displaystyle {\partial z \over \partial x} \$ come about.