Partial Derivative Simplification

BlackMelon
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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!
 
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What happens if A(x,y,z)=x2?

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