Derivative of ln(1+e^x)

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   /     x\
log\1 + E /

$$\log{\left(e^{x} + 1 \right)}$$

log(1 + E^x)

Detail solution

Let $u = e^{x} + 1$ .
The derivative of $\log{\left(u \right)}$ is $\frac{1}{u}$ .
Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(e^{x} + 1\right)$ :
1. Differentiate $e^{x} + 1$ term by term:
  1. The derivative of the constant $1$ is zero.
  2. The derivative of $e^{x}$ is itself.
  The result is: $e^{x}$
The result of the chain rule is:

$\frac{e^{x}}{e^{x} + 1}$
Now simplify:

$\frac{e^{x}}{e^{x} + 1}$

The answer is:

$\frac{e^{x}}{e^{x} + 1}$

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

   x  
  e   
------
     x
1 + E

$$\frac{e^{x}}{e^{x} + 1}$$

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

/        x         2*x \   
|     3*e       2*e    |  x
|1 - ------ + ---------|*e 
|         x           2|   
|    1 + e    /     x\ |   
\             \1 + e / /   
---------------------------
                x          
           1 + e

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

Simplify

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Derivative of ln(1+e^x)

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The solution