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Identical expressions
ln(sqrt((y- one)/(y+ one)))
ln( square root of ((y minus 1) divide by (y plus 1)))
ln( square root of ((y minus one) divide by (y plus one)))
ln(√((y-1)/(y+1)))
lnsqrty-1/y+1
ln(sqrt((y-1) divide by (y+1)))
Similar expressions
ln(sqrt((y+1)/(y+1)))
ln(sqrt((y-1)/(y-1)))

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

Derivative of ln(sqrt((y-1)/(y+1)))

The solution

You have entered [src]

   /    _______\
   |   / y - 1 |
log|  /  ----- |
   \\/   y + 1 /

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

log(sqrt((y - 1)/(y + 1)))

Detail solution

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

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

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

The answer is:

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

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

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

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

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Derivative of ln(sqrt((y-1)/(y+1)))

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The solution