Partial Derivative Homework: Prove f_{x}(tx,ty)=t^{n-1}f_{x}(x,y)

[drawar]

Homework Statement

If f is homogeneous of degree n, show that f_{x}(tx,ty)=t^{n-1}f_{x}(x,y).

Homework Equations

The Attempt at a Solution

There are many solutions out there, and here's one of them:

Since f is homogeneous of degree n, f(tx,ty)=t^{n}f(x,y) for all t, where n is a positive integer.
Taking the partial derivative wrt x
\frac{\partial }{{\partial (tx)}}f(tx,ty).\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty).\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}}
\Rightarrow t{f_x}(tx,ty) = {t^n}{f_x}(x,y) and the desired follows.

The proof is nice, but I just don't get it why from step 1 to step 2, \frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty) and then it's rewritten as {f_x}(tx,ty). Any help is very much appreciated, thanks!

 

[Fredrik]

Homework Statement

If f is homogeneous of degree n, show that f_{x}(tx,ty)=t^{n-1}f_{x}(x,y).

Homework Equations

The Attempt at a Solution

There are many solutions out there, and here's one of them:The proof is nice, but I just don't get it why from step 1 to step 2, \frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty) and then it's rewritten as {f_x}(tx,ty). Any help is very much appreciated, thanks!

It's not entirely clear to me what you're saying that the steps are (I can't make sense of what you wrote), but if you just want to apply $\partial/\partial x$ to both sides of $f(tx,ty)=t^nf(x,y)$, then you need to use the chain rule when you deal with the left-hand side.

What you need to know is that the x in the denominator of
$$\frac{\partial}{\partial x}(\text{something that involves x})$$ tells us among other things that the function that you're supposed to take a partial derivative of takes x and some other variables to the "something that involves x". For example, the x in
$$\frac{\partial}{\partial x}f(tx,ty)$$ tells us that the function that you're supposed to take a partial derivative of isn't $(x,y)\mapsto f(x,y)$ (i.e. f), but $(x,y)\mapsto f(tx,ty)$. Since $f_x$ denotes the derivative of f with respect to the first variable, we have
$$\frac{\partial}{\partial x}f(tx,ty)\neq f_x(tx,ty) =\frac{\partial}{\partial (tx)}f(tx,ty).$$ By the way, if you want all of what you're saying to be quotable, use indent tags instead of quote tags.

Like this.​

 

[HallsofIvy]

Homework Statement

If f is homogeneous of degree n, show that f_{x}(tx,ty)=t^{n-1}f_{x}(x,y).

Homework Equations

The Attempt at a Solution

There are many solutions out there, and here's one of them:


The proof is nice, but I just don't get it why from step 1 to step 2, \frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty) and then it's rewritten as {f_x}(tx,ty). Any help is very much appreciated, thanks!

I don't know what you mean by "step 1 to step 2". There is NO statement that "\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)" in what you post and it is certainly not true.

 

[drawar]

Thanks for your replies.
Sorry for not making it clear. I meant, from \frac{\partial }{{\partial (tx)}}f(tx,ty).\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty).\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}} to t{f_x}(tx,ty) = {t^n}{f_x}(x,y), doesn't it assume \frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty) or I've overlooked something?

 

[Fredrik]

Thanks for your replies.
Sorry for not making it clear. I meant, from \frac{\partial }{{\partial (tx)}}f(tx,ty).\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty).\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}} to t{f_x}(tx,ty) = {t^n}{f_x}(x,y), doesn't it assume \frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty) or I've overlooked something?

$\frac{\partial }{{\partial (tx)}}f(tx,ty)$ isn't a statement, so it can't imply anything or be implied by anything. Same thing with $\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty)$. Each step should be a statement that follows from the statement in the previous step.

 

[Fredrik]

Aaah...the periods denote multiplication. You should probably avoid that notation. OK, give me a second.

 

[Fredrik]

I have thought about it now. No, it doesn't assume that. In fact it contradicts it.

It's best to not use any symbol at all for multiplication. If you want to use a dot, use the LaTeX code \cdot.

 

[drawar]

Ok. So is it a valid proof?
I thought \frac{\partial }{{\partial (tx)}}f(tx,ty)\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty)\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}} only implies
t\frac{\partial }<br /> {{\partial (tx)}}f(tx,ty) = {t^n}\frac{{\partial f(x,y)}}<br /> {{\partial x}}?

 

[Fredrik]

Ok. So is it a valid proof?
I thought \frac{\partial }{{\partial (tx)}}f(tx,ty)\frac{{\partial (tx)}}{{\partial x}} + \frac{\partial }{{\partial (ty)}}f(tx,ty)\frac{{\partial (ty)}}{{\partial x}} = {t^n}\frac{{\partial f(x,y)}}{{\partial x}} only implies
t\frac{\partial }<br /> {{\partial (tx)}}f(tx,ty) = {t^n}\frac{{\partial f(x,y)}}<br /> {{\partial x}}?

That's right. Now you just cancel a factor of t from both sides and you're done.

 

[drawar]

That's right. Now you just cancel a factor of t from both sides and you're done.

Then \frac{\partial }<br /> {{\partial (tx)}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}<br /> {{\partial x}}, but the question asks us to prove \frac{\partial }<br /> {{\partial x}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}<br /> {{\partial x}}.

 

[Fredrik]

Then \frac{\partial }<br /> {{\partial (tx)}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}<br /> {{\partial x}}, but the question asks us to prove \frac{\partial }<br /> {{\partial x}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}<br /> {{\partial x}}.

No, it doesn't. See my first post in this thread.

 

[drawar]

No, it doesn't. See my first post in this thread.

Ahh ok. Thanks for clearing that up for me. :)