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Derivative of ctg(4x)-8*sin(2x)

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cot(4*x) - 8*sin(2*x)
8sin(2x)+cot(4x)- 8 \sin{\left(2 x \right)} + \cot{\left(4 x \right)}
cot(4*x) - 8*sin(2*x)
Detail solution
  1. Differentiate 8sin(2x)+cot(4x)- 8 \sin{\left(2 x \right)} + \cot{\left(4 x \right)} term by term:

    1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

        cot(4x)=1tan(4x)\cot{\left(4 x \right)} = \frac{1}{\tan{\left(4 x \right)}}

      2. Let u=tan(4x)u = \tan{\left(4 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(4x)\frac{d}{d x} \tan{\left(4 x \right)}:

        1. Rewrite the function to be differentiated:

          tan(4x)=sin(4x)cos(4x)\tan{\left(4 x \right)} = \frac{\sin{\left(4 x \right)}}{\cos{\left(4 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(4x)f{\left(x \right)} = \sin{\left(4 x \right)} and g(x)=cos(4x)g{\left(x \right)} = \cos{\left(4 x \right)}.

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

          1. Let u=4xu = 4 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 ddx4x\frac{d}{d x} 4 x:

            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: 44

            The result of the chain rule is:

            4cos(4x)4 \cos{\left(4 x \right)}

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

          1. Let u=4xu = 4 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 ddx4x\frac{d}{d x} 4 x:

            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: 44

            The result of the chain rule is:

            4sin(4x)- 4 \sin{\left(4 x \right)}

          Now plug in to the quotient rule:

          4sin2(4x)+4cos2(4x)cos2(4x)\frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)}}

        The result of the chain rule is:

        4sin2(4x)+4cos2(4x)cos2(4x)tan2(4x)- \frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}}

      Method #2

      1. Rewrite the function to be differentiated:

        cot(4x)=cos(4x)sin(4x)\cot{\left(4 x \right)} = \frac{\cos{\left(4 x \right)}}{\sin{\left(4 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(4x)f{\left(x \right)} = \cos{\left(4 x \right)} and g(x)=sin(4x)g{\left(x \right)} = \sin{\left(4 x \right)}.

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

        1. Let u=4xu = 4 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 ddx4x\frac{d}{d x} 4 x:

          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: 44

          The result of the chain rule is:

          4sin(4x)- 4 \sin{\left(4 x \right)}

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

        1. Let u=4xu = 4 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 ddx4x\frac{d}{d x} 4 x:

          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: 44

          The result of the chain rule is:

          4cos(4x)4 \cos{\left(4 x \right)}

        Now plug in to the quotient rule:

        4sin2(4x)4cos2(4x)sin2(4x)\frac{- 4 \sin^{2}{\left(4 x \right)} - 4 \cos^{2}{\left(4 x \right)}}{\sin^{2}{\left(4 x \right)}}

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

      1. Let u=2xu = 2 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 ddx2x\frac{d}{d x} 2 x:

        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: 22

        The result of the chain rule is:

        2cos(2x)2 \cos{\left(2 x \right)}

      So, the result is: 16cos(2x)- 16 \cos{\left(2 x \right)}

    The result is: 4sin2(4x)+4cos2(4x)cos2(4x)tan2(4x)16cos(2x)- \frac{4 \sin^{2}{\left(4 x \right)} + 4 \cos^{2}{\left(4 x \right)}}{\cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}} - 16 \cos{\left(2 x \right)}

  2. Now simplify:

    4(512sin10(x)+1280sin8(x)1152sin6(x)+448sin4(x)64sin2(x)1)cos2(4x)tan2(4x)\frac{4 \left(- 512 \sin^{10}{\left(x \right)} + 1280 \sin^{8}{\left(x \right)} - 1152 \sin^{6}{\left(x \right)} + 448 \sin^{4}{\left(x \right)} - 64 \sin^{2}{\left(x \right)} - 1\right)}{\cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}}


The answer is:

4(512sin10(x)+1280sin8(x)1152sin6(x)+448sin4(x)64sin2(x)1)cos2(4x)tan2(4x)\frac{4 \left(- 512 \sin^{10}{\left(x \right)} + 1280 \sin^{8}{\left(x \right)} - 1152 \sin^{6}{\left(x \right)} + 448 \sin^{4}{\left(x \right)} - 64 \sin^{2}{\left(x \right)} - 1\right)}{\cos^{2}{\left(4 x \right)} \tan^{2}{\left(4 x \right)}}

The graph
02468-8-6-4-2-1010-25002500
The first derivative [src]
                        2     
-4 - 16*cos(2*x) - 4*cot (4*x)
16cos(2x)4cot2(4x)4- 16 \cos{\left(2 x \right)} - 4 \cot^{2}{\left(4 x \right)} - 4
The second derivative [src]
   //       2     \                    \
32*\\1 + cot (4*x)/*cot(4*x) + sin(2*x)/
32((cot2(4x)+1)cot(4x)+sin(2x))32 \left(\left(\cot^{2}{\left(4 x \right)} + 1\right) \cot{\left(4 x \right)} + \sin{\left(2 x \right)}\right)
The third derivative [src]
   /                   2                                         \
   |    /       2     \         2      /       2     \           |
64*\- 2*\1 + cot (4*x)/  - 4*cot (4*x)*\1 + cot (4*x)/ + cos(2*x)/
64(2(cot2(4x)+1)24(cot2(4x)+1)cot2(4x)+cos(2x))64 \left(- 2 \left(\cot^{2}{\left(4 x \right)} + 1\right)^{2} - 4 \left(\cot^{2}{\left(4 x \right)} + 1\right) \cot^{2}{\left(4 x \right)} + \cos{\left(2 x \right)}\right)