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Derivative of:
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Identical expressions
log((one +x)/(one +x))
logarithm of ((1 plus x) divide by (1 plus x))
logarithm of ((one plus x) divide by (one plus x))
log1+x/1+x
log((1+x) divide by (1+x))
Similar expressions
log((1+x)/(1-x))
log((1-x)/(1+x))

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

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

The solution

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

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

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

Detail solution

Let $u = \frac{x + 1}{x + 1}$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \frac{x + 1}{x + 1}$ :
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 $g{\left(x \right)} = x + 1$ .
  
  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 $x + 1$ term by term:
    1. The derivative of the constant $1$ is zero.
    2. Apply the power rule: $x$ goes to $1$
    The result is: $1$
  Now plug in to the quotient rule:
  
  $0$
The result of the chain rule is:

$0$

The answer is:

0

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

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Simplify

The second derivative [src]

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Simplify

The third derivative [src]

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Simplify

5-я производная [src]

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Simplify