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Why p''=pp' ?

  1. Jul 27, 2009 #1
    why p''=pp' ??

    i have a function
    y(x)

    i want to substitute by variable p.

    p=y'(x)=dy/dx

    every derivative is by dx because its the smallest variable
    so why

    why dp/dy=p'


    i cant see the meaning of derivative by y

    ??
     
  2. jcsd
  3. Jul 27, 2009 #2
    Re: why p''=pp' ??

    First of all, writing dp/dy=p' implies that you're differentiating with respect to y. Otherwise, you are right, dp/dy is not equal to dp/dx in this case.

    The expression [tex]\frac{dp}{dy}[/tex] is really just a short-hand for the chain rule. The best thing to do here is not to think of p as a variable but instead as a function p(y(x))=y'(x). What it's asking is "What happens to the value of p(y(x)) as the value of y(x) changes?" That is, we want to find
    [tex]\frac{{\rm change\ in\ }p\bigl(y(x)\bigr)}{{\rm change\ in\ }y(x)},[/tex]​
    or, in "semi-short-hand" notation,
    [tex]\frac{dp\bigl(y(x)\bigr)}{dy(x)},[/tex]​
    and in standard short-hand notation,
    [tex]\frac{dp}{dy}.[/tex]​
    Now, since both p(y(x)) and y(x) are in terms of x, we can re-write these (albeit with a slight abuse of notation) as
    [tex]
    \begin{align*}
    \frac{dp}{dy} &=\frac{dp\bigl(y(x)\bigr)}{dy(x)} \\
    &= \frac{dp\bigl(y(x)\bigr)}{dy(x)} \cdot \frac{dx}{dx} \\
    &= \frac{dp\bigl(y(x)\bigr)}{dx} \biggl/ \frac{dy(x)}{dx}\\
    &= \frac{dp}{dx}\biggl/\frac{dy}{dx}\\
    &= \frac{y''(x)}{y'(x)}.
    \end{align*}
    [/tex]​
     
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