How do you differentiate #f(x)=(1-lnx)/(1+lnx)# using the quotient rule?

Question

Emily Ammerman · Accepted Answer

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

Using the quotient rule, we have: # d/dx (1-lnx)/(1+lnx) = ( (1+lnx)(d/dx (1-lnx)) - (1-lnx)(d/dx(1+lnx)) ) / (1+lnx)^2# # " " = ( (1+lnx)(-1/x) - (1-lnx)(1/x) ) / (1+lnx)^2# # " " = ( (-1/x)( 1+lnx + 1-lnx) ) / (1+lnx)^2# # " " = ( (-1/x)( 2) ) / (1+lnx)^2# # " " = - ( 2 ) / ( x(1+lnx)^2 )#

Ezekiel Alexander · Answer

undefined To differentiate $ f(x) = \frac{1 - \ln(x)}{1 + \ln(x)} $ using the quotient rule:

1. Identify $ u(x) = 1 - \ln(x) $ and $ v(x) = 1 + \ln(x) $.
2. Apply the quotient rule formula: 
$$ f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{[v(x)]^2} $$.
3. Compute the derivatives:
$$ u'(x) = \frac{d}{dx}(1 - \ln(x)) = -\frac{1}{x} $$
$$ v'(x) = \frac{d}{dx}(1 + \ln(x)) = \frac{1}{x} $$
4. Substitute the derivatives and functions into the quotient rule formula:
$$ f'(x) = \frac{(1 + \ln(x))(-\frac{1}{x}) - (1 - \ln(x))(\frac{1}{x})}{(1 + \ln(x))^2} $$.
5. Simplify the expression:
$$ f'(x) = \frac{-1 - \ln(x) + \ln(x) - \ln^2(x)}{(1 + \ln(x))^2} $$
$$ f'(x) = \frac{-1 - \ln^2(x)}{(1 + \ln(x))^2} $$.

So, the derivative of $ f(x) $ with respect to $ x $ is:
$$ f'(x) = \frac{-1 - \ln^2(x)}{(1 + \ln(x))^2} $$.