If f(x)=log(1-x)/(1+x).then what will be the value of f'(0)?

Question

Paisley Johnson · Accepted Answer

# -2#.

I presume that #log# means natural log, i.e., #ln#. #f(x)=log_e{(1-x)/(1+x)}=ln{(1-x)/(1+x)}#. Using the usual rules of #log#, we have, #f(x)=ln(1-x)-ln(1+x)# #rArr f'(x)=d/dx{ln(1-x)}-d/dx{ln(1+x)}#, #=1/(1-x)*d/dx(1-x)-1/(1+x)*d/dx(1+x)...............[because," the Chain Rule],"# #=1/(1-x)*(-1)-1/(1+x)*1#, #=1/(x-1)-1/(x+1)#, #={(x+1)-(x-1)}/{(x+1)(x-1)}#, #:. f'(x)=2/(x^2-1)#. #rArr f'(0)=2/(0-1)=-2#.

Christopher Allers · Answer

undefined To find the value of $ f'(0) $, we first need to differentiate the function $ f(x) $ with respect to $ x $ using the quotient rule. Then, we substitute $ x = 0 $ into the derivative expression.

Given $ f(x) = \frac{{\log(1-x)}}{{1+x}} $,

Using the quotient rule, the derivative of $ f(x) $ with respect to $ x $ is:

$$ f'(x) = \frac{{(1+x) \cdot \frac{{d}}{{dx}}(\log(1-x)) - \log(1-x) \cdot \frac{{d}}{{dx}}(1+x)}}{{(1+x)^2}} $$

$$ f'(x) = \frac{{(1+x) \cdot \left(\frac{{-1}}{{1-x}}\right) - \log(1-x) \cdot 1}}{{(1+x)^2}} $$

$$ f'(x) = \frac{{-1-x - \log(1-x)}}{{(1+x)(1-x)}} $$

Now, substitute $ x = 0 $ into $ f'(x) $:

$$ f'(0) = \frac{{-1 - 0 - \log(1-0)}}{{(1+0)(1-0)}} $$

$$ f'(0) = \frac{{-1 - \log(1)}}{{1}} $$

$$ f'(0) = \frac{{-1 - 0}}{{1}} $$

$$ f'(0) = -1 $$