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SahanaC Group Title

Can someone walk me step by step through the proof of the product rule? I didn't really understand how the professor got from u(x + deltax)v(x+ deltax) - u(x)v(x) to (u(x+ deltax) - u(x))v(x + deltax) + u(x)v(x + deltax) - v(x).

  • one year ago
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  1. zigs Group Title
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    We start with: \[u(x+\Delta x)v(x+\Delta x) - u(x)v(x)\] Now to try to get the equation in a form that we would like to see, let's add the following: \[u(x)v(x+\Delta x) - u(x)v(x+\Delta x)\] We are adding and subtracting the same two terms, so effectively we are adding zero to our original equation which doesn't alter it at all. So, we wind up with: \[u(x+\Delta x)v(x+\Delta x) - u(x)v(x+\Delta x) + u(x)v(x+\Delta x)-u(x)v(x)\] Grouping terms, we come up with: \[\left\{ u(x+\Delta x)-u(x) \right\} v(x+\Delta x) + u(x)\left\{ v(x+\Delta x)-v(x) \right\}\] Which is the same result that the professor derived. He didn't go through the above steps explicitly, but basically this is how he came up with the result.

    • one year ago
  2. SahanaC Group Title
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    Why exactly did you add the u(x)v(x+Δx)−u(x)v(x+Δx) ? I can kinda understand the part after that.. but I was just confused as to why (and where) that part came from.

    • one year ago
  3. creeksider Group Title
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    I expect you understand why you're ALLOWED to do this, but just in case you didn't catch that point, the reason is that we're adding and subtracting the same thing, and that is the same as adding zero. The more interesting question is what conceivable thought process would lead someone to come up with this particular approach. The answer lies in our desire to get something in a form that resembles the numerator of the difference quotient:\[f(x+\Delta x)-f(x)\]because when we get something looking like that we can combine it with the Delta x in the denominator to construct a derivative. Here's a schematic of what's going on. In this schematic, the variables a, b, c and d stand for the four elements of the original equation. For example, a stands for u(x+deltax). We start here:\[ab-cd\]Then we add and subtract the same number, which is the product of one factor of the first term and one factor of the second one, as indicated in parens:\[ab+(-ac+ac)-cd\]Then we change the grouping:\[(ab-ac)+(ac-cd)\]And then we factor:\[a(b-c)+c(a-d)\]When performed on the actual expression, this process puts us in position to complete the proof because (b - c) and (a - d) match the numerator for difference quotients that will allow us to identify derivatives. While you didn't ask about the rest of the proof, it's worth noting that one piece that might appear insignificant is actually crucial. Later in the proof we have to use the fact that \[\lim_{\Delta x \rightarrow 0} u(x+\Delta x)=u(x)\]It may seem obvious that this has to be true, but we're allowed to do this for a specific reason. We're told at the outset the u and v are differentiable, and in an earlier lecture Professor Jerison gave us a proof that when a function is differentiable it is continuous. These are the details that connect the dots for this necessary part of the proof.

    • one year ago
  4. SahanaC Group Title
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    Thank you so much! That is extremely helpful!

    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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