Whats the derivative of $ln((1+x)/(1-x))$ ?

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Answer 1 · 2016-11-06T03:33:46

$dy/dx= 2/(1 - x^2)$

Let $y = lnu$ and $u = (1 + x)/(1 - x)$ .

We will use the chain rule to differentiate this problem. However, we must first find the derivative of each function.

$y' = 1/u$

$u' = (1(1 - x) - (-1(1 + x)))/(1 - x)^2$

$u' = (1 - x - (-1 - x))/(1 - x)^2$

$u' = (1 - x + 1 + x)/(1- x)^2$

$u' = 2/(1 - x)^2$

Now, by the chain rule:

$dy/dx= 1/u xx 2/(1 - x)^2$

$dy/dx= 1/((1 + x)/(1 - x)) xx 2/(1 - x)^2$

$dy/dx = (1 - x)/(1 + x) xx 2/(1 - x)^2$

$dy/dx= 2/((1 + x)(1 - x))$

$dy/dx= 2/(1 - x^2)$

Hopefully this helps!

Whats the derivative of ln((1+x)/(1-x))ln((1+x)/(1-x))?