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

Apr 10, 2017

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

#### Explanation:

Assuming 'log' = $\ln$

$y = \frac{1 + \ln x}{1 - \ln x}$

Applying the Quotient rule and the standard differential for $\ln x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \ln x\right) \left(\frac{1}{x}\right) - \left(1 + \ln x\right) \left(- \frac{1}{x}\right)}{{\left(1 - \ln x\right)}^{2}}$

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

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

Apr 10, 2017

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{\log {x}^{2}}{\left(x \ln 10\right) {\left(1 - \log x\right)}^{2}}$

#### Explanation:

$\text{Assuming } \log x = {\log}_{10} x$

$\text{Then } \frac{d}{\mathrm{dx}} \left({\log}_{10} x\right) = \frac{1}{x \ln 10}$

differentiate using the $\textcolor{b l u e}{\text{quotient rule}}$

$\text{Given " y=(g(x))/(h(x))" then}$

$\textcolor{red}{\overline{\underline{| \textcolor{w h i t e}{\frac{2}{2}} \textcolor{b l a c k}{\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{h \left(x\right) g ' \left(x\right) - g \left(x\right) h ' \left(x\right)}{h \left(x\right)} ^ 2} \textcolor{w h i t e}{\frac{2}{2}} |}}}$

$\text{here } g \left(x\right) = 1 + \log x \Rightarrow g ' \left(x\right) = \frac{1}{x \ln 10}$

$\text{and } h \left(x\right) = 1 - \log x \Rightarrow h ' \left(x\right) = \frac{1}{x \ln 10}$

$\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\left(1 - \log x\right) \left(\frac{1}{x \ln 10}\right) - \left(1 + \log x\right) \left(\frac{1}{x \ln 10}\right)}{1 - \log x} ^ 2$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = \frac{\frac{1}{x \ln 10} \left(1 - \log x - 1 - \log x\right)}{1 - \log x} ^ 2$

$\textcolor{w h i t e}{\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}} = - \frac{\log {x}^{2}}{\left(x \ln 10\right) {\left(1 - \log x\right)}^{2}}$