# How do you differentiate f(t)= sqrt(( 1+ ln(t) ) / ( 1 - ln(t) ) ?

$f ' \left(t\right) = \setminus \frac{2}{\left(1 - \setminus \ln t\right) \setminus \sqrt{1 - {\left(\setminus \ln t\right)}^{2}}}$

#### Explanation:

Given function:

$f \left(t\right) = \setminus \sqrt{\setminus \frac{1 + \setminus \ln t}{1 - \setminus \ln t}}$

$f \left(t\right) = \setminus \frac{\setminus \sqrt{1 + \setminus \ln t}}{\setminus \sqrt{1 - \setminus \ln t}}$

Differentiating above function w.r.t. $t$ using quotient rule as follows

$f ' \left(t\right) = \setminus \frac{\setminus \sqrt{1 - \setminus \ln t} \frac{d}{\mathrm{dt}} \setminus \sqrt{1 + \setminus \ln t} - \setminus \sqrt{1 + \setminus \ln t} \frac{d}{\mathrm{dt}} \setminus \sqrt{1 - \setminus \ln t}}{{\left(\setminus \sqrt{1 - \setminus \ln t}\right)}^{2}}$

$= \setminus \frac{\setminus \sqrt{1 - \setminus \ln t} \frac{1}{2 t \setminus \sqrt{1 + \setminus \ln t}} - \setminus \sqrt{1 + \setminus \ln t} \frac{- 1}{2 t \setminus \sqrt{1 - \setminus \ln t}}}{1 - \setminus \ln t}$

$= \setminus \frac{1 - \setminus \ln t + 1 + \setminus \ln t}{\left(1 - \setminus \ln t\right) \setminus \sqrt{1 - {\left(\setminus \ln t\right)}^{2}}}$

$= \setminus \frac{2}{\left(1 - \setminus \ln t\right) \setminus \sqrt{1 - {\left(\setminus \ln t\right)}^{2}}}$