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dudeperuvian Group Title

help me find dy if if y=xlnx then dy=

  • one year ago
  • one year ago

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  1. vf321 Group Title
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    Product rule. \[f(x)=x\]\[g(x)=ln(x)\]\[y=f(x)g(x)\]\[dy=(f'(x)g(x)+f(x)g'(x))dx\]

    • one year ago
  2. mtaOS Group Title
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    Yes It seems you will have to use the integration by parts. @vf321 is correctr.

    • one year ago
  3. mtaOS Group Title
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    Did you understand ?

    • one year ago
  4. dudeperuvian Group Title
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    hmm

    • one year ago
  5. dudeperuvian Group Title
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    so it would be (x+lnx)dx ? :3

    • one year ago
  6. vf321 Group Title
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    NO! I made up some functions, f(x) and g(x). Can you tell me what they are? (Hint: Look up). Then find f'(x) and g'(x).

    • one year ago
  7. dudeperuvian Group Title
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    or 1 + lnx dx

    • one year ago
  8. mtaOS Group Title
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    You have to choose which one you integerate and which one you derive.

    • one year ago
  9. dudeperuvian Group Title
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    1lnx+x(...)

    • one year ago
  10. vf321 Group Title
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    yes. what is the ...?

    • one year ago
  11. dudeperuvian Group Title
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    (x+lnx)dx ?

    • one year ago
  12. vf321 Group Title
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    no... Let's look at it one at a time. What is f(x)?

    • one year ago
  13. dudeperuvian Group Title
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    f(x)= x so f' is equal to one. i get that, but how to derive lnx ;S

    • one year ago
  14. dudeperuvian Group Title
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    thats how i got 1lnx+x(...)

    • one year ago
  15. vf321 Group Title
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    That's a definition. d/dx(lnx) = 1/x

    • one year ago
  16. vf321 Group Title
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    so now plug into the product rule: \[dy = (f(x)g'(x) + f'(x)g(x))dx\]

    • one year ago
  17. dudeperuvian Group Title
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    \[ \frac{ lnx+x }{ x}\]

    • one year ago
  18. dudeperuvian Group Title
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    forgot the dx

    • one year ago
  19. vf321 Group Title
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    dy = (1+lnx)dx

    • one year ago
  20. dudeperuvian Group Title
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    why?!?!?! -_-

    • one year ago
  21. PhoenixFire Group Title
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    \[f(x) = x\] \[f^{\prime}(x) = 1\] \[g(x) = \ln(x)\] \[g^{\prime}(x) = {1 \over x}\] Product rule: \[{{\delta fg}\over{\delta y}} = f^{\prime}g + fg^{\prime}\] Now substitute the equations calculated above. \[{\delta y \over \delta x} = 1\ln(x) + x{1 \over x}\] \[{\delta y \over \delta x} = \ln(x) + 1\] the x * 1/x is the same as x/x, which is the same as 1.

    • one year ago
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