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anonymous

  • 5 years ago

What is the integral of y/x dx?

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  1. amistre64
    • 5 years ago
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    we can change this a bit: dy = (y/x) dx or dy/y = dx/x and integrate (S) both sides. (S) (1/y) dy = (S) (1/x) dx ln(y) = ln(x) + C e^(ln(y)) = e^(ln(x) + C) y=e^(ln(x)) * e^C y=Cx This is my best guess :)

  2. Gina
    • 5 years ago
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    yes you right

  3. amistre64
    • 5 years ago
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    yay!! :)

  4. anonymous
    • 5 years ago
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    Well…really, \[\int\limits(y/x)dx\]means to integrate with respect to x, treating everything else as a constant. Thus, \[y \int\limits(1/x)=y \ln \left| x \right|+C.\]Just sayin'.

  5. amistre64
    • 5 years ago
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    I would love to agree with you carly; but, ... If we undo that we get: d(f(x))/dx = d(y ln(x))/dx f'(x) = y*(1/x)*(dx/dx) + (dydx)*(ln(x)) f'(x) = y/x + y' ln(x). ; which is not the same as (y/x) Right?

  6. anonymous
    • 5 years ago
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    What is the derivative of ln x? Now what's that times a constant y, hmm?

  7. amistre64
    • 5 years ago
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    But the "y" is not a constant; it is an implicit function of "x". If "y" is an independant variable of f(x,y) then yes, we would hold "y" as a constant to find the rate of change of f(x,y) with respect to "x".

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