Christos
  • Christos
Logarithmic differantiation, y = (x^3 -2x)^ln(x) I got a very long result I dont know if my solution is correct
Mathematics
schrodinger
  • schrodinger
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Christos
  • Christos
y'=(x^3 - 2x)^ln(x)lnx^3-2x+lnx*(2x-2)/(x^3-2x)
Christos
  • Christos
I forgot an 1/x next to (x^3 - 2x)
Christos
  • Christos
multiplication

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.Sam.
  • .Sam.
I got\[\Large y'=[\frac{(3x^2-2)\ln(x)}{x^3-2x}+\frac{\ln(x^3-2x)}{x}](x^3-2x)^{\ln(x)}\]
Christos
  • Christos
hm
reemii
  • reemii
only thing to correct in your answer: it's \(3x^2-2\) instead of \(2x-2\), and you forgot some parentheses.
Christos
  • Christos
Yea I just noticed that error, and the parenthesis I know, I just surpassed them. Other than that all ok ??
reemii
  • reemii
yes
.Sam.
  • .Sam.
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 \[\frac{1}{y}y'=\ln(x)\frac{1}{x^3-2x}(3x^2-2)+\ln(x^3-2x)(\frac{1}{x})\]
Christos
  • Christos
thank you guys

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