Whats the derivative of y=1/2 ln[(1+x)/(1-x)]?

1 Answer
Sep 3, 2017

1/(1-x^2).

Explanation:

y=1/2ln[(1+x)/(1-x)].

Using the Familiar Rules of log. fun., we get,

y=1/2[ln(1+x)-ln(1-x)].

:. dy/dx=1/2[d/dx{ln(1+x)}-d/dx{ln(1-x)}],

=1/2[1/(1+x)*d/dx(1+x)-1/(1-x)*d/dx(1-x)]...[because," the Chain Rule"],

=1/2[1/(1+x)(1)-1/(1-x)(-1)],

=1/2[1/(1+x)+1/(1-x)],

=1/2[{(1+x)+(1-x)}/{(1+x)(1-x)}],

=1/cancel2[cancel2/(1-x^2)].

:. dy/dx=1/(1-x^2), is the desired Derivative.

Enjoy Maths.!