Derivative of ln(1-e^x)

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   /     x\
log\1 - E /

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

log(1 - E^x)

Detail solution

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

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

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

The answer is:

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

The first derivative [src]

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

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

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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