Given the limit: $$\lim _{\Delta t\to 0}\frac{\Delta x\Delta y}{\Delta t}$$ where both x and y are function of t. Is this equivalent to derivative of product or something similar?
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what exactly are $\Delta x$ and $\Delta y$? – user251257 Aug 10 '15 at 14:12
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Does it matter? If $$\lim _{\Delta t\to 0}\frac{\Delta x}{\Delta t}$$ was asked, would you ask for what is \Delta x – Salihcyilmaz Aug 10 '15 at 14:14
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it is important to understand your formula, isn't it? – user251257 Aug 10 '15 at 14:15
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We just know x and y are t dependent. Not explicitly given. – Salihcyilmaz Aug 10 '15 at 14:16
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I meant, is $\Delta x = x(\Delta t) - x(0)$ or something like that? – user251257 Aug 10 '15 at 14:17
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you can think like $$x_k - x_(k-1)$$ – Salihcyilmaz Aug 10 '15 at 14:21
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Is there a special reason why the expression is like this ? – callculus42 Aug 10 '15 at 14:52
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sorry for the format error in last comment. – user251257 Aug 10 '15 at 15:30
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@Salihcyilmaz To answer your question. It is not equivalent to the derivative of a product of two functions. I see just now, that Zach has already answered the question. – callculus42 Aug 10 '15 at 16:03
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Good question, it indeed is not equivalent to taking the product of two derivatives. But, as you are about to see, your intuition was fairly close.
$$\lim_{\Delta t \to 0} {{\Delta x \Delta y} \over {\Delta t}}=\lim_{\Delta t \to 0} {{\Delta x} \over {\sqrt{\Delta t}}} \cdot \lim_{\Delta t \to 0} {{ \Delta y} \over {\sqrt{\Delta t}}}={{dx} \over {dt^{1/2}}} \cdot {{dy} \over {dt^{1/2}}}$$
So the result is the product of two Alpha Derivatives. In short, the alpha derivative "smooths over" areas of a function with a near infinite rate of change. So, for well behaved functions,
$$\lim_{\Delta t \to 0} {{\Delta x \Delta y} \over {\Delta t}}=0$$
Zach466920
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