Mister Exam

Derivative of ctg(x/3)

Function f() - derivative -N order at the point
v

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The solution

You have entered [src]
   /x\
cot|-|
   \3/
cot(x3)\cot{\left(\frac{x}{3} \right)}
cot(x/3)
Detail solution
  1. There are multiple ways to do this derivative.

    Method #1

    1. Rewrite the function to be differentiated:

      cot(x3)=1tan(x3)\cot{\left(\frac{x}{3} \right)} = \frac{1}{\tan{\left(\frac{x}{3} \right)}}

    2. Let u=tan(x3)u = \tan{\left(\frac{x}{3} \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(x3)\frac{d}{d x} \tan{\left(\frac{x}{3} \right)}:

      1. Rewrite the function to be differentiated:

        tan(x3)=sin(x3)cos(x3)\tan{\left(\frac{x}{3} \right)} = \frac{\sin{\left(\frac{x}{3} \right)}}{\cos{\left(\frac{x}{3} \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(x3)f{\left(x \right)} = \sin{\left(\frac{x}{3} \right)} and g(x)=cos(x3)g{\left(x \right)} = \cos{\left(\frac{x}{3} \right)}.

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

        1. Let u=x3u = \frac{x}{3}.

        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 ddxx3\frac{d}{d x} \frac{x}{3}:

          1. The derivative of a constant times a function is the constant times the derivative of the function.

            1. Apply the power rule: xx goes to 11

            So, the result is: 13\frac{1}{3}

          The result of the chain rule is:

          cos(x3)3\frac{\cos{\left(\frac{x}{3} \right)}}{3}

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

        1. Let u=x3u = \frac{x}{3}.

        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 ddxx3\frac{d}{d x} \frac{x}{3}:

          1. The derivative of a constant times a function is the constant times the derivative of the function.

            1. Apply the power rule: xx goes to 11

            So, the result is: 13\frac{1}{3}

          The result of the chain rule is:

          sin(x3)3- \frac{\sin{\left(\frac{x}{3} \right)}}{3}

        Now plug in to the quotient rule:

        sin2(x3)3+cos2(x3)3cos2(x3)\frac{\frac{\sin^{2}{\left(\frac{x}{3} \right)}}{3} + \frac{\cos^{2}{\left(\frac{x}{3} \right)}}{3}}{\cos^{2}{\left(\frac{x}{3} \right)}}

      The result of the chain rule is:

      sin2(x3)3+cos2(x3)3cos2(x3)tan2(x3)- \frac{\frac{\sin^{2}{\left(\frac{x}{3} \right)}}{3} + \frac{\cos^{2}{\left(\frac{x}{3} \right)}}{3}}{\cos^{2}{\left(\frac{x}{3} \right)} \tan^{2}{\left(\frac{x}{3} \right)}}

    Method #2

    1. Rewrite the function to be differentiated:

      cot(x3)=cos(x3)sin(x3)\cot{\left(\frac{x}{3} \right)} = \frac{\cos{\left(\frac{x}{3} \right)}}{\sin{\left(\frac{x}{3} \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(x3)f{\left(x \right)} = \cos{\left(\frac{x}{3} \right)} and g(x)=sin(x3)g{\left(x \right)} = \sin{\left(\frac{x}{3} \right)}.

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

      1. Let u=x3u = \frac{x}{3}.

      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 ddxx3\frac{d}{d x} \frac{x}{3}:

        1. The derivative of a constant times a function is the constant times the derivative of the function.

          1. Apply the power rule: xx goes to 11

          So, the result is: 13\frac{1}{3}

        The result of the chain rule is:

        sin(x3)3- \frac{\sin{\left(\frac{x}{3} \right)}}{3}

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

      1. Let u=x3u = \frac{x}{3}.

      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 ddxx3\frac{d}{d x} \frac{x}{3}:

        1. The derivative of a constant times a function is the constant times the derivative of the function.

          1. Apply the power rule: xx goes to 11

          So, the result is: 13\frac{1}{3}

        The result of the chain rule is:

        cos(x3)3\frac{\cos{\left(\frac{x}{3} \right)}}{3}

      Now plug in to the quotient rule:

      sin2(x3)3cos2(x3)3sin2(x3)\frac{- \frac{\sin^{2}{\left(\frac{x}{3} \right)}}{3} - \frac{\cos^{2}{\left(\frac{x}{3} \right)}}{3}}{\sin^{2}{\left(\frac{x}{3} \right)}}

  2. Now simplify:

    13sin2(x3)- \frac{1}{3 \sin^{2}{\left(\frac{x}{3} \right)}}


The answer is:

13sin2(x3)- \frac{1}{3 \sin^{2}{\left(\frac{x}{3} \right)}}

The graph
02468-8-6-4-2-1010-50005000
The first derivative [src]
         2/x\
      cot |-|
  1       \3/
- - - -------
  3      3   
cot2(x3)313- \frac{\cot^{2}{\left(\frac{x}{3} \right)}}{3} - \frac{1}{3}
The second derivative [src]
  /       2/x\\    /x\
2*|1 + cot |-||*cot|-|
  \        \3//    \3/
----------------------
          9           
2(cot2(x3)+1)cot(x3)9\frac{2 \left(\cot^{2}{\left(\frac{x}{3} \right)} + 1\right) \cot{\left(\frac{x}{3} \right)}}{9}
The third derivative [src]
   /       2/x\\ /         2/x\\
-2*|1 + cot |-||*|1 + 3*cot |-||
   \        \3// \          \3//
--------------------------------
               27               
2(cot2(x3)+1)(3cot2(x3)+1)27- \frac{2 \left(\cot^{2}{\left(\frac{x}{3} \right)} + 1\right) \left(3 \cot^{2}{\left(\frac{x}{3} \right)} + 1\right)}{27}
The graph
Derivative of ctg(x/3)