- #1

Apashanka

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- Thread starter Apashanka
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- #1

Apashanka

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- #2

Mark44

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Not quite. Above, you're using the definition of the (partial) derivative of the product of two functions, which is a limit.

Corrected, this would be $$\lim_{h \to 0}

\frac{u(x+h, y)v(x+h, y) - u(x, y)v(x,y )}{h}$$

- #3

Apashanka

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Ok using ##\frac{\partial}{\partial x}u(x,y)v(x,y)=\frac{\partial u(x,y)}{\partial x}v(x,y)+u(x,y)\frac{\partial v(x,y)}{\partial x}## can't it be ##\frac{u(x+dx,y)-u(x,y)}{dx}v(x,y)+u(x,y)\frac{v(x+dx,y)-v(x,y)}{dx}??##Not quite. Above, you're using the definition of the (partial) derivative of the product of two functions, which is a limit.

Corrected, this would be $$\lim_{h \to 0}

\frac{u(x+h, y)v(x+h, y) - u(x, y)v(x,y )}{h}$$

- #4

Mark44

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Again, not quite -- you are trying to use the definitions of the two partial derivatives without including that these are limits.Ok using ##\frac{\partial}{\partial x}u(x,y)v(x,y)=\frac{\partial u(x,y)}{\partial x}v(x,y)+u(x,y)\frac{\partial v(x,y)}{\partial x}## can't it be ##\frac{u(x+dx,y)-u(x,y)}{dx}v(x,y)+u(x,y)\frac{v(x+dx,y)-v(x,y)}{dx}??##

The corrected version would be $$\lim_{h \to 0}\left(\frac{u(x+h,y)-u(x,y)}{h}\right) v(x,y)+u(x,y)\lim_{h \to 0}\left(\frac{v(x+h,y)-v(x,y)}{h}\right)$$

- #5

Apashanka

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So which is correct post 2 or post 4Again, not quite -- you are trying to use the definitions of the two partial derivatives without including that these are limits.

The corrected version would be $$\lim_{h \to 0}\left(\frac{u(x+h,y)-u(x,y)}{h}\right) v(x,y)+u(x,y)\lim_{h \to 0}\left(\frac{v(x+h,y)-v(x,y)}{h}\right)$$

- #6

Mark44

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Both are correct. It's possible to derive what I wrote in post 4 from what is in post 2.So which is correct post 2 or post 4

- #7

HallsofIvy

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