What is the first differential of #y = (1+logx)/(1-logx)#?

2 Answers
Alan N.
Apr 10, 2017

#dy/dx= 2/(x(1-lnx)^2#

Explanation:

Assuming 'log' = #ln#

#y = (1+lnx)/(1-lnx)#

Applying the Quotient rule and the standard differential for #lnx#

#dy/dx = ((1-lnx)(1/x) - (1+lnx)(-1/x)) / ((1-lnx)^2)#

#= (1-lnx+1 +lnx)/ (x(1-lnx)^2#

#= 2/(x(1-lnx)^2#

Jim G.
Apr 10, 2017

#dy/dx=-(logx^2)/((xln10)(1-logx)^2)#

Explanation:

#"Assuming " log x=log_(10) x#

#"Then " d/dx(log_(10)x)=1/(xln10)#

differentiate using the #color(blue)"quotient rule"#

#"Given " y=(g(x))/(h(x))" then"#

#color(red)(bar(ul(|color(white)(2/2)color(black)(dy/dx=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))#

#"here " g(x)=1+logxrArrg'(x)=1/(xln10)#

#"and " h(x)=1-logxrArrh'(x)=1/(xln10)#

#rArrdy/dx=((1-logx)(1/(xln10))-(1+logx)(1/(xln10)))/(1-logx)^2#

#color(white)(rArrdy/dx)=(1/(xln10)(1-logx-1-logx))/(1-logx)^2#

#color(white)(rArrdy/dx)=-(logx^2)/((xln10)(1-logx)^2)#

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