Whats the derivative of y=1/2 ln[(1+x)/(1-x)]? Calculus Differentiating Logarithmic Functions Differentiating Logarithmic Functions with Base e 1 Answer Ratnaker Mehta 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.! Answer link Related questions What is the derivative of f(x)=ln(g(x)) ? What is the derivative of f(x)=ln(x^2+x) ? What is the derivative of f(x)=ln(e^x+3) ? What is the derivative of f(x)=x*ln(x) ? What is the derivative of f(x)=e^(4x)*ln(1-x) ? What is the derivative of f(x)=ln(x)/x ? What is the derivative of f(x)=ln(cos(x)) ? What is the derivative of f(x)=ln(tan(x)) ? What is the derivative of f(x)=sqrt(1+ln(x) ? What is the derivative of f(x)=(ln(x))^2 ? See all questions in Differentiating Logarithmic Functions with Base e Impact of this question 6854 views around the world You can reuse this answer Creative Commons License