Question

Differentiate the following functions with respect to x :

$\frac{1+logx}{1-logx}$

Answer 1

We have,

$\frac{d}{dx}\left(\frac{1+logx}{1-logx}\right)$

using quotient rule,

$\frac{(1-logx)\frac{d}{dx}(1+logx)-(1+logx)\frac{d}{dx}\left(10logx\right)}{(1-logx{)}^2}$

$=\frac{(1-logx)\times \frac{1}{x}-(1+logx)(-\frac{1}{x})}{(1-logx{)}^2}=\frac{1-logx+1+logx}{x(1-logx{)}^2}=\frac{2}{x(1-logx{)}^2}$