Partial Derivative Simplification

[BlackMelon]

Homework Statement

A is a function of x, y, and z. Simplify:
$$\frac{\partial^2A}{\partial x^2}+\frac{\partial^2A}{\partial y^2}$$

Relevant Equations

Simplify $$\frac{\partial^2A}{\partial x^2}+\frac{\partial^2A}{\partial y^2}$$

Hi there!

I would like to know if the following simplification is correct or not:
Let A be a function of x, y, and z

$$\frac{\partial^2A}{\partial x^2}+\frac{\partial^2A}{\partial y^2}$$
$$=\ \frac{\partial^2A\partial y^2+\partial^2A\partial x^2}{\partial x^2\partial y^2}$$
$$=\frac{\partial^2A\left(\partial y^2+\partial x^2\right)}{\partial x^2\partial y^2}$$
$$=\ \frac{\partial^2A\left(\partial z^2\right)}{\partial x^2\partial y^2}$$
$$=0\ \ \left(since\frac{\partial z^2}{\partial y^2}=0\right)$$

Thanks!

 

[Frabjous]

What happens if A(x,y,z)=x²?

There has to be more to the problem than what you state.

 

[pasmith]

Homework Statement: A is a function of x, y, and z. Simplify:
$$\frac{\partial^2A}{\partial x^2}+\frac{\partial^2A}{\partial y^2}$$
Relevant Equations: Simplify $$\frac{\partial^2A}{\partial x^2}+\frac{\partial^2A}{\partial y^2}$$

Hi there!

I would like to know if the following simplification is correct or not:
Let A be a function of x, y, and z

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

\dfrac{\partial^2 A}{\partial x^2} is not a fraction. \dfrac{\partial^2}{\partial x^2} is the second partial derivative operator with respect to x, keeping other variables constant. It does not have a "numerator" or "denominator" which you can manipulate separately in order to put a sum of such operators over a "common denominator". Without knowing more about A, there is nothing about the expression <br /> \frac{\partial^2 A}{\partial x^2} + \frac{\partial^2 A}{\partial y^2} that can be simplified.

 

[hutchphd]

Thanks!

Was this a formal assignment? (from where?)
Please do not miss the larger point (from @Frabjous) that a good way to check a proposed result is to try to create a counterexample. You only need one.....that is what makes the scientific method so powerful.