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Integral of d{x}:
log((1+x)/(1-x))
Limit of the function:
log((1+x)/(1-x))
Identical expressions
log((one +x)/(one -x))
logarithm of ((1 plus x) divide by (1 minus x))
logarithm of ((one plus x) divide by (one minus x))
log1+x/1-x
log((1+x) divide by (1-x))
Similar expressions
log((1+x)/(1+x))
log(((1+x)/(1-x))^(1/8))/log(3)
log((1-x)/(1-x))

The derivative of the function/ log((1+x)/(1-x))

Derivative of log((1+x)/(1-x))

The solution

You have entered [src]

   /1 + x\
log|-----|
   \1 - x/

$$\log{\left(\frac{x + 1}{1 - x} \right)}$$

log((1 + x)/(1 - x))

Detail solution

Let $u = \frac{x + 1}{1 - x}$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \frac{x + 1}{1 - x}$ :
1. Apply the quotient rule, which is:
  
  $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
  
  $f{\left(x \right)} = x + 1$ and $g{\left(x \right)} = 1 - x$ .
  
  To find $\frac{d}{d x} f{\left(x \right)}$ :
  1. Differentiate $x + 1$ term by term:
    1. The derivative of the constant $1$ is zero.
    2. Apply the power rule: $x$ goes to $1$
    The result is: $1$
  To find $\frac{d}{d x} g{\left(x \right)}$ :
  1. Differentiate $1 - x$ term by term:
    1. The derivative of the constant $1$ is zero.
    2. The derivative of a constant times a function is the constant times the derivative of the function.
      1. Apply the power rule: $x$ goes to $1$
      So, the result is: $-1$
    The result is: $-1$
  Now plug in to the quotient rule:
  
  $\frac{2}{\left(1 - x\right)^{2}}$
The result of the chain rule is:

$\frac{2}{\left(1 - x\right) \left(x + 1\right)}$
Now simplify:

$- \frac{2}{x^{2} - 1}$

The answer is:

$- \frac{2}{x^{2} - 1}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

        /  1      1 + x  \
(1 - x)*|----- + --------|
        |1 - x          2|
        \        (1 - x) /
--------------------------
          1 + x

$$\frac{\left(1 - x\right) \left(\frac{1}{1 - x} + \frac{x + 1}{\left(1 - x\right)^{2}}\right)}{x + 1}$$

Simplify

The second derivative [src]

/    1 + x \ /    1       1   \
|1 - ------|*|- ----- - ------|
\    -1 + x/ \  1 + x   -1 + x/
-------------------------------
             1 + x

$$\frac{\left(1 - \frac{x + 1}{x - 1}\right) \left(- \frac{1}{x + 1} - \frac{1}{x - 1}\right)}{x + 1}$$

Simplify

The third derivative [src]

  /    1 + x \ /   1           1              1        \
2*|1 - ------|*|-------- + --------- + ----------------|
  \    -1 + x/ |       2           2   (1 + x)*(-1 + x)|
               \(1 + x)    (-1 + x)                    /
--------------------------------------------------------
                         1 + x

$$\frac{2 \left(1 - \frac{x + 1}{x - 1}\right) \left(\frac{1}{\left(x + 1\right)^{2}} + \frac{1}{\left(x - 1\right) \left(x + 1\right)} + \frac{1}{\left(x - 1\right)^{2}}\right)}{x + 1}$$

Simplify

The graph

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Derivative of log((1+x)/(1-x))

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The solution