Multivariable Calculus Equality involving the Difference Quotient

Question

We have

$-(M(x,y+\Delta{y})-M(x,y))\Delta{y} = -(\frac{ \partial M }{\partial y} *\Delta x) \Delta y = -\Delta y (M(x,y+ \Delta y) + M(x,y))$

.

Expanding:

$ -(\frac{ \partial M }{\partial y}*\Delta x) \Delta y = \frac{-M(x,y+\Delta y)\Delta x \Delta y+ M(x,y) \Delta x \Delta y}{\Delta y} ={-M(x,y+\Delta y)\Delta x + M(x,y) \Delta x} = -\Delta x (M(x,y+\Delta y)\Delta x + M(x,y)) $

I don't see how

$-(M(x,y+\Delta{y})-M(x,y))\Delta{y} = -(\frac{ \partial M }{\partial y} *\Delta x) \Delta y$

Because if that were true then $ {-M(x,y+\Delta y)\Delta x + M(x,y) \Delta x} = -(M(x,y+\Delta{y})-M(x,y))\Delta{y} $ Which is not true unless $ \Delta x = \Delta y$

We don't know that $ \Delta x = \Delta y$. It can be the case that $ \Delta x \neq \Delta y$ .

I can't post an image and the context of the problem comes from this link:

The story starts here

Image 1

Image 2

Image 3 is just supplementary useless information

So if you looked at image two, you'd see the equation I highlighted in yellow for the Top and Bottom Part.

Answer 1

You've misread the statement. It's actually

$$ \big(M(x,y+\Delta y) - M(x,y)\big) \Delta x = \frac{\partial M}{\partial y}\Delta y \Delta x $$

Since

$$ \frac{\partial M}{\partial y} \approx \frac{M(x,y+\Delta y) - M(x,y)}{\Delta y}$$