Question

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

Answer 1

`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`

Answer 2

`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)`