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

The solution

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

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

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

Detail solution

Apply the product rule:

$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$

$f{\left(x \right)} = x$ ; to find $\frac{d}{d x} f{\left(x \right)}$ :
1. Apply the power rule: $x$ goes to $1$
$g{\left(x \right)} = \log{\left(\frac{x + 1}{1 - x} \right)}$ ; to find $\frac{d}{d x} g{\left(x \right)}$ :
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:
  
  $\frac{2}{\left(1 - x\right) \left(x + 1\right)}$
The result is: $\frac{2 x}{\left(1 - x\right) \left(x + 1\right)} + \log{\left(\frac{x + 1}{1 - x} \right)}$
Now simplify:

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

The answer is:

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

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

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

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

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

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Piecewise:

The solution