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Identical expressions
y=lnsqrt((two x+ one)/(2-x))
y equally ln square root of ((2x plus 1) divide by (2 minus x))
y equally ln square root of ((two x plus one) divide by (2 minus x))
y=ln√((2x+1)/(2-x))
y=lnsqrt2x+1/2-x
y=lnsqrt((2x+1) divide by (2-x))
Similar expressions
y=lnsqrt((2x-1)/(2-x))
y=lnsqrt((2x+1)/(2+x))

The derivative of the function/ y=lnsqrt((2x+1)/(2-x))

Derivative of y=lnsqrt((2x+1)/(2-x))

The solution

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

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

log(sqrt((2*x + 1)/(2 - x)))

Detail solution

Let $u = \sqrt{\frac{2 x + 1}{2 - x}}$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \sqrt{\frac{2 x + 1}{2 - x}}$ :
1. Let $u = \frac{2 x + 1}{2 - x}$ .
2. Apply the power rule: $\sqrt{u}$ goes to $\frac{1}{2 \sqrt{u}}$
3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \frac{2 x + 1}{2 - x}$ :
  1. Apply the quotient rule, which is:
    
    $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
    
    $f{\left(x \right)} = 2 x + 1$ and $g{\left(x \right)} = 2 - x$ .
    
    To find $\frac{d}{d x} f{\left(x \right)}$ :
    1. Differentiate $2 x + 1$ 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: $2$
      The result is: $2$
    To find $\frac{d}{d x} g{\left(x \right)}$ :
    1. Differentiate $2 - x$ term by term:
      1. The derivative of the constant $2$ 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:
    
    $\frac{5}{\left(2 - x\right)^{2}}$
  The result of the chain rule is:
  
  $\frac{5}{2 \sqrt{\frac{2 x + 1}{2 - x}} \left(2 - x\right)^{2}}$
The result of the chain rule is:

$\frac{5 \frac{2 - x}{2 x + 1}}{2 \left(2 - x\right)^{2}}$
Now simplify:

$- \frac{5}{2 \left(x - 2\right) \left(2 x + 1\right)}$

The answer is:

$- \frac{5}{2 \left(x - 2\right) \left(2 x + 1\right)}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

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

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

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

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

Simplify

The graph

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Derivative of y=lnsqrt((2x+1)/(2-x))

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The solution