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Identical expressions
ln((one +x)/(one -x))-(one /x)
ln((1 plus x) divide by (1 minus x)) minus (1 divide by x)
ln((one plus x) divide by (one minus x)) minus (one divide by x)
ln1+x/1-x-1/x
ln((1+x) divide by (1-x))-(1 divide by x)
Similar expressions
ln((1+x)/(1-x))+(1/x)
ln((1+x)/(1+x))-(1/x)
ln((1-x)/(1-x))-(1/x)

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

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

The solution

You have entered [src]

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

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

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

Detail solution

Differentiate $\log{\left(\frac{x + 1}{1 - x} \right)} - \frac{1}{x}$ term by term:
1. Let $u = \frac{x + 1}{1 - x}$ .
2. The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
3. 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.
        
        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:
  
  $\frac{2}{\left(1 - x\right) \left(x + 1\right)}$
4. The derivative of a constant times a function is the constant times the derivative of the function.
  1. Apply the power rule: $\frac{1}{x}$ goes to $- \frac{1}{x^{2}}$
  So, the result is: $\frac{1}{x^{2}}$
The result is: $\frac{2}{\left(1 - x\right) \left(x + 1\right)} + \frac{1}{x^{2}}$
Now simplify:

$- \frac{x^{2} + 1}{x^{4} - x^{2}}$

The answer is:

$- \frac{x^{2} + 1}{x^{4} - x^{2}}$

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

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

Simplify

The second derivative [src]

           1 + x           1 + x    
       1 - ------      1 - ------   
  2        -1 + x          -1 + x   
- -- - ---------- - ----------------
   3           2    (1 + x)*(-1 + x)
  x     (1 + x)

$$- \frac{1 - \frac{x + 1}{x - 1}}{\left(x + 1\right)^{2}} - \frac{1 - \frac{x + 1}{x - 1}}{\left(x - 1\right) \left(x + 1\right)} - \frac{2}{x^{3}}$$

Simplify

The third derivative [src]

  /         1 + x            1 + x               1 + x    \
  |     1 - ------       1 - ------          1 - ------   |
  |3        -1 + x           -1 + x              -1 + x   |
2*|-- + ---------- + ----------------- + -----------------|
  | 4           3                    2          2         |
  \x     (1 + x)     (1 + x)*(-1 + x)    (1 + x) *(-1 + x)/

$$2 \left(\frac{1 - \frac{x + 1}{x - 1}}{\left(x + 1\right)^{3}} + \frac{1 - \frac{x + 1}{x - 1}}{\left(x - 1\right) \left(x + 1\right)^{2}} + \frac{1 - \frac{x + 1}{x - 1}}{\left(x - 1\right)^{2} \left(x + 1\right)} + \frac{3}{x^{4}}\right)$$

Simplify

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

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Piecewise:

The solution