amnahali2565 amnahali2565
  • 26-03-2018
  • Mathematics
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Find the derivative of y with respect to x. y= ((lnx)/(2+lnx))

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PiPhi
PiPhi PiPhi
  • 26-03-2018
Using the known theorem [tex]\left(\frac{f}{g}\right)'=\frac{f'g-g'f}{g^2}[/tex] you get
[tex]y'=\left(\frac{\ln x}{2+\ln x}\right) '=\frac{(\ln x)'(2+\ln x)-(2+\ln x)'\ln x}{(2+\ln x)^2}=\frac{\frac{1}{x}(2+\ln x)-\frac{1}{x}\ln x}{(2+\ln x)^2}[/tex].

Therefore, [tex]y'=\frac{\frac{2+\ln x}{x}-\frac{\ln x}{x}}{(2+\ln x)^2}=\frac{\frac{2}{x}}{(2+\ln x)^2}=\frac{2}{x(2+\ln x)^2}[/tex].
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