Rearranging

**Simplicity** · November 18th 2010, 11:52 AM

What does the rearranging lead to?

$\frac{\partial a}{\partial x} \partial y = \frac{\partial b}{\partial y} \partial x$

Dividing through by $\partial x$ , which would it lead to?

(1)
$\frac{\partial a}{\partial x^2} \partial y = \frac{\partial b}{\partial y}$

(2)
$\frac{\partial ^2 a}{\partial x^2} \partial y = \frac{\partial b}{\partial y}$

Equation (1) or (2)?

**Soroban** · November 18th 2010, 01:00 PM

Hello, Simplicity!

What does the rearranging lead to?

. . $\displaystyle \frac{\partial a}{\partial x} \partial y \:=\: \frac{\partial b}{\partial y} \partial x$

Dividing through by $\partial x$ , which would it lead to?

$\displaystyle (1)\;\frac{\partial a}{\partial x^2} \partial y \:=\: \frac{\partial b}{\partial y}$
. . . . . . . or
$\displaystyle (2)\;\frac{\partial ^2 a}{\partial x^2} \partial y \:=\: \frac{\partial b}{\partial y}$

We have: . $\displaystyle \frac{\partial a}{\partial x}\partial y \;=\;\frac{\partial b}{\partial y}\partial x$

Dividing through by $\partial x$ , we have: . $\displaystyle \frac{\partial a}{\partial x}\cdot \frac{\partial y}{\partial x} \;=\;\frac{\partial b}{\partial y}\cdot\frac{\partial x}{\partial x}$

Then we have: . $\displaystyle\frac{\partial a}{\partial x}\cdot\frac{\partial y}{\partial x} \;=\;\frac{\partial b}{\partial y}$

. . which can be written: . $\displaystyle \frac{\partial a\partial y}{\partial x^2} \;=\;\frac{\partial b}{\partial y}$ . . . answer (1)

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