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Christos

Logarithmic differantiation, y = (x^3 -2x)^ln(x) I got a very long result I dont know if my solution is correct

  • 10 months ago
  • 10 months ago

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  1. Christos
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    y'=(x^3 - 2x)^ln(x)lnx^3-2x+lnx*(2x-2)/(x^3-2x)

    • 10 months ago
  2. Christos
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    I forgot an 1/x next to (x^3 - 2x)

    • 10 months ago
  3. Christos
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    multiplication

    • 10 months ago
  4. .Sam.
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    I got\[\Large y'=[\frac{(3x^2-2)\ln(x)}{x^3-2x}+\frac{\ln(x^3-2x)}{x}](x^3-2x)^{\ln(x)}\]

    • 10 months ago
  5. Christos
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    hm

    • 10 months ago
  6. reemii
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    only thing to correct in your answer: it's \(3x^2-2\) instead of \(2x-2\), and you forgot some parentheses.

    • 10 months ago
  7. Christos
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    Yea I just noticed that error, and the parenthesis I know, I just surpassed them. Other than that all ok ??

    • 10 months ago
  8. reemii
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    yes

    • 10 months ago
  9. .Sam.
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    Yeah \[y=(x^3-2x)^{\ln(x)} \\ \\ \ln(y)=\ln(x^3-2x)^{\ln(x)} \\ \\ \ln(y)=\ln(x)\ln(x^3-2x)\] Implicit differentiation , and product rule + chain rule for RHS \[\frac{1}{y}y'=\ln(x)\frac{1}{x^3-2x}(3x^2-2)+\ln(x^3-2x)(\frac{1}{x})\]

    • 10 months ago
  10. Christos
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    thank you guys

    • 10 months ago
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