Derivative of cot(1/x)

The solution

You have entered [src]

   /  1\
cot|1*-|
   \  x/

$$\cot{\left(1 \cdot \frac{1}{x} \right)}$$

d /   /  1\\
--|cot|1*-||
dx\   \  x//

$$\frac{d}{d x} \cot{\left(1 \cdot \frac{1}{x} \right)}$$

Transform

Detail solution

There are multiple ways to do this derivative.
Method #1
1. Rewrite the function to be differentiated:
  
  $\cot{\left(1 \cdot \frac{1}{x} \right)} = \frac{1}{\tan{\left(1 \cdot \frac{1}{x} \right)}}$
2. Let $u = \tan{\left(1 \cdot \frac{1}{x} \right)}$ .
3. Apply the power rule: $\frac{1}{u}$ goes to $- \frac{1}{u^{2}}$
4. Then, apply the chain rule. Multiply by $\frac{d}{d x} \tan{\left(1 \cdot \frac{1}{x} \right)}$ :
  1. Rewrite the function to be differentiated:
    
    $\tan{\left(1 \cdot \frac{1}{x} \right)} = \frac{\sin{\left(1 \cdot \frac{1}{x} \right)}}{\cos{\left(1 \cdot \frac{1}{x} \right)}}$
  2. Apply the quotient rule, which is:
    
    $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
    
    $f{\left(x \right)} = \sin{\left(1 \cdot \frac{1}{x} \right)}$ and $g{\left(x \right)} = \cos{\left(1 \cdot \frac{1}{x} \right)}$ .
    
    To find $\frac{d}{d x} f{\left(x \right)}$ :
    1. Let $u = 1 \cdot \frac{1}{x}$ .
    2. The derivative of sine is cosine:
      
      $\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$
    3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 1 \cdot \frac{1}{x}$ :
      
      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:
      
      $- \frac{1}{x^{2}}$
      
      The result of the chain rule is:
      
      $- \frac{\cos{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}$
    To find $\frac{d}{d x} g{\left(x \right)}$ :
    1. Let $u = 1 \cdot \frac{1}{x}$ .
    2. The derivative of cosine is negative sine:
      
      $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
    3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 1 \cdot \frac{1}{x}$ :
      
      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:
      
      $- \frac{1}{x^{2}}$
      
      The result of the chain rule is:
      
      $\frac{\sin{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}$
    Now plug in to the quotient rule:
    
    $\frac{- \frac{\sin^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}} - \frac{\cos^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}}{\cos^{2}{\left(1 \cdot \frac{1}{x} \right)}}$
  The result of the chain rule is:
  
  $- \frac{- \frac{\sin^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}} - \frac{\cos^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}}{\cos^{2}{\left(1 \cdot \frac{1}{x} \right)} \tan^{2}{\left(1 \cdot \frac{1}{x} \right)}}$
Method #2
1. Rewrite the function to be differentiated:
  
  $\cot{\left(1 \cdot \frac{1}{x} \right)} = \frac{\cos{\left(1 \cdot \frac{1}{x} \right)}}{\sin{\left(1 \cdot \frac{1}{x} \right)}}$
2. Apply the quotient rule, which is:
  
  $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
  
  $f{\left(x \right)} = \cos{\left(1 \cdot \frac{1}{x} \right)}$ and $g{\left(x \right)} = \sin{\left(1 \cdot \frac{1}{x} \right)}$ .
  
  To find $\frac{d}{d x} f{\left(x \right)}$ :
  1. Let $u = 1 \cdot \frac{1}{x}$ .
  2. The derivative of cosine is negative sine:
    
    $\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}$
  3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 1 \cdot \frac{1}{x}$ :
    1. Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = 1$ 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:
      
      $- \frac{1}{x^{2}}$
    The result of the chain rule is:
    
    $\frac{\sin{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}$
  To find $\frac{d}{d x} g{\left(x \right)}$ :
  1. Let $u = 1 \cdot \frac{1}{x}$ .
  2. The derivative of sine is cosine:
    
    $\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$
  3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 1 \cdot \frac{1}{x}$ :
    1. Apply the quotient rule, which is:
      
      $\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$
      
      $f{\left(x \right)} = 1$ 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:
      
      $- \frac{1}{x^{2}}$
    The result of the chain rule is:
    
    $- \frac{\cos{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}$
  Now plug in to the quotient rule:
  
  $\frac{\frac{\sin^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}} + \frac{\cos^{2}{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}}{\sin^{2}{\left(1 \cdot \frac{1}{x} \right)}}$
Now simplify:

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

The answer is:

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

The graph

Plot the graph f(x)

Plot the graph f'(x)

The first derivative [src]

 /        2/  1\\ 
-|-1 - cot |1*-|| 
 \         \  x// 
------------------
         2        
        x

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

Simplify

The second derivative [src]

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

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

Simplify

The third derivative [src]

                /           2/1\        /1\        2/1\\
                |    1 + cot |-|   6*cot|-|   2*cot |-||
  /       2/1\\ |            \x/        \x/         \x/|
2*|1 + cot |-||*|3 + ----------- - -------- + ---------|
  \        \x// |          2          x            2   |
                \         x                       x    /
--------------------------------------------------------
                            4                           
                           x

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

Simplify

The graph

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Identical expressions

Similar expressions

Derivative of cot(1/x)

The graph:

Piecewise:

The solution

Method #1

Method #2