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-(e^(one /x- one))/((x- one)^ two)
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Similar expressions
(e^(1/x-1))/((x-1)^2)
-(e^(1/x+1))/((x-1)^2)
-(e^(1/x-1))/((x+1)^2)
What you mean?
(-e^(1/(x - 1*1)))/((x - 1*1)^2)
(-e^(1/x - 1*1))/((x - 1*1)^2)
(-e^(1/x - 1*1))/((x - 1*1)^2)

The derivative of the function/ -(e^(1/x-1))/((x-1)^2)

You entered:

-(e^(1/x-1))/((x-1)^2)

What you mean?

(-e^(1/(x - 1*1)))/((x - 1*1)^2)

(-e^(1/x - 1*1))/((x - 1*1)^2)

(-e^(1/x - 1*1))/((x - 1*1)^2)

Derivative of -(e^(1/x-1))/((x-1)^2)

The solution

You have entered [src]

    1     
  1*- - 1 
    x     
-e        
----------
        2 
 (x - 1)

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

  /    1     \
  |  1*- - 1 |
  |    x     |
d |-e        |
--|----------|
dx|        2 |
  \ (x - 1)  /

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

Transform

Detail solution

Apply the quotient rule, which is:

$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$

$f{\left(x \right)} = - e^{\left(-1\right) 1 + 1 \cdot \frac{1}{x}}$ and $g{\left(x \right)} = \left(x - 1\right)^{2}$ .

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

$\frac{\left(2 x - 2\right) e^{\left(-1\right) 1 + 1 \cdot \frac{1}{x}} + \frac{\left(x - 1\right)^{2} e^{\left(-1\right) 1 + 1 \cdot \frac{1}{x}}}{x^{2}}}{\left(x - 1\right)^{4}}$
Now simplify:

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

The answer is:

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

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

     1                     1    
   1*- - 1               1*- - 1
     x                     x    
  e           (2 - 2*x)*e       
----------- - ------------------
 2        2               4     
x *(x - 1)         (x - 1)

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

Simplify

The second derivative [src]

 /                1              \       1 
 |            2 + -              |  -1 + - 
 |    6           x        4     |       x 
-|--------- + ----- + -----------|*e       
 |        2      3     2         |         
 \(-1 + x)      x     x *(-1 + x)/         
-------------------------------------------
                         2                 
                 (-1 + x)

$$- \frac{\left(\frac{6}{\left(x - 1\right)^{2}} + \frac{2 + \frac{1}{x}}{x^{3}} + \frac{4}{x^{2} \left(x - 1\right)}\right) e^{-1 + \frac{1}{x}}}{\left(x - 1\right)^{2}}$$

Simplify

The third derivative [src]

/                1    6                             \        
|            6 + -- + -                     /    1\ |       1
|                 2   x                   6*|2 + -| |  -1 + -
|    24          x             18           \    x/ |       x
|--------- + ---------- + ------------ + -----------|*e      
|        3        4        2         2    3         |        
\(-1 + x)        x        x *(-1 + x)    x *(-1 + x)/        
-------------------------------------------------------------
                                  2                          
                          (-1 + x)

$$\frac{\left(\frac{24}{\left(x - 1\right)^{3}} + \frac{6 \cdot \left(2 + \frac{1}{x}\right)}{x^{3} \left(x - 1\right)} + \frac{6 + \frac{6}{x} + \frac{1}{x^{2}}}{x^{4}} + \frac{18}{x^{2} \left(x - 1\right)^{2}}\right) e^{-1 + \frac{1}{x}}}{\left(x - 1\right)^{2}}$$

Simplify

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What you mean?

Derivative of -(e^(1/x-1))/((x-1)^2)

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