Mo6rganLillualhothe Mo6rganLillualhothe
  • 21-11-2016
  • Mathematics
contestada

trying to find derivative of y = x^(lnx), show step-by-step solution

Respuesta :

alsteva alsteva
  • 24-11-2016
[tex]y=x^{\ln{x}} \\ \ln{y}=\ln{x^{\ln{x}}} \\ \ln{y}=\ln{x}\ln{x} \\ \frac{1}{y} dy= (\frac{1}{x} \ln{x}+ \frac{1}{x} \ln{x})dx \\ \\\frac{1}{y} y'= \frac{2}{x} \ln{x} \\ \\y'=y\frac{2}{x} \ln{x} \\ \\y'=x^{\ln{x}}\frac{2}{x} \ln{x} \\ \\y'= \frac{x^{\ln{x}}2\ln{x}}{x} \\ \\y'=2x^{\ln{x}-1}\ln{x}[/tex]
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