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-7*cot(2*x)

Derivative of -7*cot(2*x)

Function f() - derivative -N order at the point
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-7*cot(2*x)
7cot(2x)- 7 \cot{\left(2 x \right)}
d              
--(-7*cot(2*x))
dx             
ddx(7cot(2x))\frac{d}{d x} \left(- 7 \cot{\left(2 x \right)}\right)
Detail solution
  1. The derivative of a constant times a function is the constant times the derivative of the function.

    1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

        cot(2x)=1tan(2x)\cot{\left(2 x \right)} = \frac{1}{\tan{\left(2 x \right)}}

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

        1. Rewrite the function to be differentiated:

          tan(2x)=sin(2x)cos(2x)\tan{\left(2 x \right)} = \frac{\sin{\left(2 x \right)}}{\cos{\left(2 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(2x)f{\left(x \right)} = \sin{\left(2 x \right)} and g(x)=cos(2x)g{\left(x \right)} = \cos{\left(2 x \right)}.

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

          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)}

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

          1. Let u=2xu = 2 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 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:

            2sin(2x)- 2 \sin{\left(2 x \right)}

          Now plug in to the quotient rule:

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

        The result of the chain rule is:

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

      Method #2

      1. Rewrite the function to be differentiated:

        cot(2x)=cos(2x)sin(2x)\cot{\left(2 x \right)} = \frac{\cos{\left(2 x \right)}}{\sin{\left(2 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(2x)f{\left(x \right)} = \cos{\left(2 x \right)} and g(x)=sin(2x)g{\left(x \right)} = \sin{\left(2 x \right)}.

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

        1. Let u=2xu = 2 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 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:

          2sin(2x)- 2 \sin{\left(2 x \right)}

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

        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)}

        Now plug in to the quotient rule:

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

    So, the result is: 7(2sin2(2x)+2cos2(2x))cos2(2x)tan2(2x)\frac{7 \cdot \left(2 \sin^{2}{\left(2 x \right)} + 2 \cos^{2}{\left(2 x \right)}\right)}{\cos^{2}{\left(2 x \right)} \tan^{2}{\left(2 x \right)}}

  2. Now simplify:

    281cos(4x)\frac{28}{1 - \cos{\left(4 x \right)}}


The answer is:

281cos(4x)\frac{28}{1 - \cos{\left(4 x \right)}}

The graph
02468-8-6-4-2-1010-5000050000
The first derivative [src]
           2     
14 + 14*cot (2*x)
14cot2(2x)+1414 \cot^{2}{\left(2 x \right)} + 14
The second derivative [src]
    /       2     \         
-56*\1 + cot (2*x)/*cot(2*x)
56(cot2(2x)+1)cot(2x)- 56 \left(\cot^{2}{\left(2 x \right)} + 1\right) \cot{\left(2 x \right)}
The third derivative [src]
    /       2     \ /         2     \
112*\1 + cot (2*x)/*\1 + 3*cot (2*x)/
112(cot2(2x)+1)(3cot2(2x)+1)112 \left(\cot^{2}{\left(2 x \right)} + 1\right) \left(3 \cot^{2}{\left(2 x \right)} + 1\right)
The graph
Derivative of -7*cot(2*x)