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cot(1/x)
The derivative of the function/ cot(1/x)

Derivative of cot(1/x)

Function f() - derivative -N order at the point
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You have entered [src] 
   /  1\
cot|1*-|
   \  x/
cot⁡(1⋅1x)\cot{\left(1 \cdot \frac{1}{x} \right)} 
d /   /  1\\
--|cot|1*-||
dx\   \  x//
ddxcot⁡(1⋅1x)\frac{d}{d x} \cot{\left(1 \cdot \frac{1}{x} \right)}
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Detail solution

  1. There are multiple ways to do this derivative.

    Method #1

    1. Rewrite the function to be differentiated:

      cot⁡(1⋅1x)=1tan⁡(1⋅1x)\cot{\left(1 \cdot \frac{1}{x} \right)} = \frac{1}{\tan{\left(1 \cdot \frac{1}{x} \right)}}

    2. Let u=tan⁡(1⋅1x)u = \tan{\left(1 \cdot \frac{1}{x} \right)}.

    3. Apply the power rule: 1u\frac{1}{u} goes to −1u2- \frac{1}{u^{2}}

    4. Then, apply the chain rule. Multiply by ddxtan⁡(1⋅1x)\frac{d}{d x} \tan{\left(1 \cdot \frac{1}{x} \right)}:

      1. Rewrite the function to be differentiated:

        tan⁡(1⋅1x)=sin⁡(1⋅1x)cos⁡(1⋅1x)\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:

        ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=sin⁡(1⋅1x)f{\left(x \right)} = \sin{\left(1 \cdot \frac{1}{x} \right)} and g(x)=cos⁡(1⋅1x)g{\left(x \right)} = \cos{\left(1 \cdot \frac{1}{x} \right)}.

        To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

        1. Let u=1⋅1xu = 1 \cdot \frac{1}{x}.

        2. The derivative of sine is cosine:

          ddusin⁡(u)=cos⁡(u)\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}

        3. Then, apply the chain rule. Multiply by ddx1⋅1x\frac{d}{d x} 1 \cdot \frac{1}{x}:

          1. Apply the quotient rule, which is:

            ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=1f{\left(x \right)} = 1 and g(x)=xg{\left(x \right)} = x.

            To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

            1. The derivative of the constant 11 is zero.

            To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

            1. Apply the power rule: xx goes to 11

            Now plug in to the quotient rule:

            −1x2- \frac{1}{x^{2}}

          The result of the chain rule is:

          −cos⁡(1⋅1x)x2- \frac{\cos{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}

        To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

        1. Let u=1⋅1xu = 1 \cdot \frac{1}{x}.

        2. The derivative of cosine is negative sine:

          dducos⁡(u)=−sin⁡(u)\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}

        3. Then, apply the chain rule. Multiply by ddx1⋅1x\frac{d}{d x} 1 \cdot \frac{1}{x}:

          1. Apply the quotient rule, which is:

            ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=1f{\left(x \right)} = 1 and g(x)=xg{\left(x \right)} = x.

            To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

            1. The derivative of the constant 11 is zero.

            To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

            1. Apply the power rule: xx goes to 11

            Now plug in to the quotient rule:

            −1x2- \frac{1}{x^{2}}

          The result of the chain rule is:

          sin⁡(1⋅1x)x2\frac{\sin{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}

        Now plug in to the quotient rule:

        −sin⁡2(1⋅1x)x2−cos⁡2(1⋅1x)x2cos⁡2(1⋅1x)\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:

      −−sin⁡2(1⋅1x)x2−cos⁡2(1⋅1x)x2cos⁡2(1⋅1x)tan⁡2(1⋅1x)- \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⁡(1⋅1x)=cos⁡(1⋅1x)sin⁡(1⋅1x)\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:

      ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=cos⁡(1⋅1x)f{\left(x \right)} = \cos{\left(1 \cdot \frac{1}{x} \right)} and g(x)=sin⁡(1⋅1x)g{\left(x \right)} = \sin{\left(1 \cdot \frac{1}{x} \right)}.

      To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

      1. Let u=1⋅1xu = 1 \cdot \frac{1}{x}.

      2. The derivative of cosine is negative sine:

        dducos⁡(u)=−sin⁡(u)\frac{d}{d u} \cos{\left(u \right)} = - \sin{\left(u \right)}

      3. Then, apply the chain rule. Multiply by ddx1⋅1x\frac{d}{d x} 1 \cdot \frac{1}{x}:

        1. Apply the quotient rule, which is:

          ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=1f{\left(x \right)} = 1 and g(x)=xg{\left(x \right)} = x.

          To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

          1. The derivative of the constant 11 is zero.

          To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

          1. Apply the power rule: xx goes to 11

          Now plug in to the quotient rule:

          −1x2- \frac{1}{x^{2}}

        The result of the chain rule is:

        sin⁡(1⋅1x)x2\frac{\sin{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}

      To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

      1. Let u=1⋅1xu = 1 \cdot \frac{1}{x}.

      2. The derivative of sine is cosine:

        ddusin⁡(u)=cos⁡(u)\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}

      3. Then, apply the chain rule. Multiply by ddx1⋅1x\frac{d}{d x} 1 \cdot \frac{1}{x}:

        1. Apply the quotient rule, which is:

          ddxf(x)g(x)=−f(x)ddxg(x)+g(x)ddxf(x)g2(x)\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(x)=1f{\left(x \right)} = 1 and g(x)=xg{\left(x \right)} = x.

          To find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

          1. The derivative of the constant 11 is zero.

          To find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

          1. Apply the power rule: xx goes to 11

          Now plug in to the quotient rule:

          −1x2- \frac{1}{x^{2}}

        The result of the chain rule is:

        −cos⁡(1⋅1x)x2- \frac{\cos{\left(1 \cdot \frac{1}{x} \right)}}{x^{2}}

      Now plug in to the quotient rule:

      sin⁡2(1⋅1x)x2+cos⁡2(1⋅1x)x2sin⁡2(1⋅1x)\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)}}

  2. Now simplify:

    1x2sin⁡2(1x)\frac{1}{x^{2} \sin^{2}{\left(\frac{1}{x} \right)}}

The answer is:

1x2sin⁡2(1x)\frac{1}{x^{2} \sin^{2}{\left(\frac{1}{x} \right)}}

The graph
02468-8-6-4-2-1010-500500
Plot the graph f(x)
Plot the graph f'(x)
The first derivative [src] 
 /        2/  1\\ 
-|-1 - cot |1*-|| 
 \         \  x// 
------------------
         2        
        x
−−cot⁡2(1⋅1x)−1x2- \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
2(−1+cot⁡(1x)x)(cot⁡2(1x)+1)x3\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
2(cot⁡2(1x)+1)(3−6cot⁡(1x)x+2cot⁡2(1x)x2+cot⁡2(1x)+1x2)x4\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
Derivative of cot(1/x)
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