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

In summary, we are given the equation f(tx,ty)=t^n*f(x,y) for a homogeneous function of degree n. Using the chain rule, we can show that t*f_x(tx,ty)=t^n*f_x(x,y). Simplifying, we get f_x(tx,ty)=t^(n-1)*f_x(x,y), which is what we were asked to prove.
  • #1
drawar
132
0

Homework Statement



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

Homework Equations


The Attempt at a Solution



There are many solutions out there, and here's one of them:
Since [itex]f[/itex] is homogeneous of degree [itex]n[/itex], [itex]f(tx,ty)=t^{n}f(x,y)[/itex] for all [itex]t[/itex], where [itex]n[/itex] is a positive integer.
Taking the partial derivative wrt [itex]x[/itex]
[itex]\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}}[/itex]
[itex] \Rightarrow t{f_x}(tx,ty) = {t^n}{f_x}(x,y)[/itex] and the desired follows.

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

Homework Statement



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

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, [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] and then it's rewritten as [itex]{f_x}(tx,ty)[/itex]. 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.​
 
Last edited:
  • #3
drawar said:

Homework Statement



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

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, [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] and then it's rewritten as [itex]{f_x}(tx,ty)[/itex]. 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 "[itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex]" in what you post and it is certainly not true.
 
  • #4
Thanks for your replies.
Sorry for not making it clear. I meant, from [itex]\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}}[/itex] to [itex]t{f_x}(tx,ty) = {t^n}{f_x}(x,y)[/itex], doesn't it assume [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] or I've overlooked something?
 
  • #5
drawar said:
Thanks for your replies.
Sorry for not making it clear. I meant, from [itex]\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}}[/itex] to [itex]t{f_x}(tx,ty) = {t^n}{f_x}(x,y)[/itex], doesn't it assume [itex]\frac{\partial }{{\partial (tx)}}f(tx,ty) = \frac{\partial }{{\partial x}}f(tx,ty)[/itex] 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.
 
  • #6
Aaah...the periods denote multiplication. You should probably avoid that notation. :smile: OK, give me a second.
 
  • #7
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.
 
  • #8
Ok. So is it a valid proof?
I thought [itex]\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}}[/itex] only implies
[itex]t\frac{\partial }
{{\partial (tx)}}f(tx,ty) = {t^n}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex]?
 
  • #9
drawar said:
Ok. So is it a valid proof?
I thought [itex]\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}}[/itex] only implies
[itex]t\frac{\partial }
{{\partial (tx)}}f(tx,ty) = {t^n}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex]?
That's right. Now you just cancel a factor of t from both sides and you're done.
 
  • #10
Fredrik said:
That's right. Now you just cancel a factor of t from both sides and you're done.

Then [itex]\frac{\partial }
{{\partial (tx)}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex], but the question asks us to prove [itex]\frac{\partial }
{{\partial x}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex].
 
  • #11
drawar said:
Then [itex]\frac{\partial }
{{\partial (tx)}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex], but the question asks us to prove [itex]\frac{\partial }
{{\partial x}}f(tx,ty) = {t^{n - 1}}\frac{{\partial f(x,y)}}
{{\partial x}}[/itex].
No, it doesn't. See my first post in this thread.
 
  • #12
Fredrik said:
No, it doesn't. See my first post in this thread.

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

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

1. What is a partial derivative?

A partial derivative is a type of derivative used in multivariate calculus to determine the rate of change of a function with respect to one of its variables, while holding all other variables constant.

2. How do I prove a partial derivative homework problem?

To prove a partial derivative problem, you will need to use the definition of a partial derivative and the properties of derivatives. Start by taking the partial derivative of the function with respect to the given variable, and then substitute the given values into the resulting expression. If the final expression is equal to the given derivative, then the proof is complete.

3. What does f_{x}(tx,ty) mean in the given problem?

f_{x}(tx,ty) represents the partial derivative of the function f with respect to the variable x, where the values of x and y are multiplied by a constant t. This represents a change in the x-direction while holding the y-direction constant.

4. Why is t^{n-1} included in the given problem?

The t^{n-1} term is included to account for the chain rule when taking the partial derivative with respect to tx or ty. This term helps to simplify the derivative expression and make it easier to prove.

5. Can I use the power rule when proving this partial derivative problem?

Yes, the power rule can be applied when proving a partial derivative problem. However, you will also need to use the chain rule and other properties of derivatives to fully prove the given equation.

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