How Does the Chain Rule Apply to Homogeneous Functions in Calculus?

[anatta]

Homework Statement

A function is called homogeneous of degree n if it satisfied the equation f(tx,ty) =t^(n) f(x,y), for all t, where n is a positive integer and f has continuous 2nd order partial derivatives.

If f is homogeneous of degree n, show that df/dx (tx,ty) = t^(n-1) df/dx(x,y)

*Using df/dx for partial derviatives.

The Attempt at a Solution

Basically I've taken the partial derivatives of each side of the definition of homogeneous equation above, applying the chain rule (that's what section the problem is in). What I get is:

d/d(tx) (tx,ty) * t = t^n d/dx(x,y)

which simplifies to df/d(tx) (tx,ty) = t^(n-1) df/dx(x,y).

It feels like I'm done, but I can't figure out-conceptually or otherwise- why the partial derivative with respect to (tx) of f(tx,ty) is the same as the partial derivative with respect to x of f(tx,ty).

Thanks in advance for help!

 

[HallsofIvy]

Homework Statement

A function is called homogeneous of degree n if it satisfied the equation f(tx,ty) =t^(n) f(x,y), for all t, where n is a positive integer and f has continuous 2nd order partial derivatives.

If f is homogeneous of degree n, show that df/dx (tx,ty) = t^(n-1) df/dx(x,y)

*Using df/dx for partial derviatives.

The Attempt at a Solution

Basically I've taken the partial derivatives of each side of the definition of homogeneous equation above, applying the chain rule (that's what section the problem is in). What I get is:

d/d(tx) (tx,ty) * t = t^n d/dx(x,y)

which simplifies to df/d(tx) (tx,ty) = t^(n-1) df/dx(x,y).

It feels like I'm done, but I can't figure out-conceptually or otherwise- why the partial derivative with respect to (tx) of f(tx,ty) is the same as the partial derivative with respect to x of f(tx,ty).

It's not! $\partial f(tx,ty)/\partial (tx)$ is the same as $\partial f(u,ty)/\partial u$ for any variable u. $\partial f(tx,ty)/\partial x= \partial f(xt,ty)/\partial (xt) \partial f(tx)\partial x$.

In other words $\partial f(tx,ty)/\partial x$ is t times $\partial f(tx,ty)/\partial (tx)$.

Thanks in advance for help!

 

[anatta]

Yes, that's what I got when I applied the chain rule, although maybe my notation was confusing.

d(tx,ty)/d(tx) times t = t^n d(x,y)/dx
(the right side of the equation is the partial derivative with wrt x of the function given in the original definition)

once you divide by t you get

d(tx,ty)/d(tx) =t^(n-1) d(x,y)/dx

What I don't get is how to deal with the partial derivative w.r.t. (tx). The left side of the equation needs to be d(tx,ty) w.r.t to x for the requested proof.

 

[Defennder]

If f is homogeneous of degree n, show that df/dx (tx,ty) = t^(n-1) df/dx(x,y)

Is this even true? Try it out for the case $$f(x,y) \ = \ xy^3 + x^2 y^2.$$

Did you mean to say $$t^n \frac{\partial}{\partial x} f(x,y)$$ on the RHS instead? I assume t is an arbitrary constant.

 

[anatta]

You're right, it doesn't seem to work for the case you give. That is definitely how the problem is stated, though. It's from Stewart Calculus, 5ed, 14.6, #55. I guess it could be a typo...?