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Christos

  • 2 years ago

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

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

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

  3. Christos
    • 2 years ago
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    multiplication

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

  5. Christos
    • 2 years ago
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    hm

  6. reemii
    • 2 years ago
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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.

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

  8. reemii
    • 2 years ago
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    yes

  9. .Sam.
    • 2 years ago
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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. Christos
    • 2 years ago
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    thank you guys

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