Mister Exam

Derivative of y=ctg9x

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

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The solution

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cot(9*x)
cot(9x)\cot{\left(9 x \right)}
cot(9*x)
Detail solution
  1. There are multiple ways to do this derivative.

    Method #1

    1. Rewrite the function to be differentiated:

      cot(9x)=1tan(9x)\cot{\left(9 x \right)} = \frac{1}{\tan{\left(9 x \right)}}

    2. Let u=tan(9x)u = \tan{\left(9 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(9x)\frac{d}{d x} \tan{\left(9 x \right)}:

      1. Rewrite the function to be differentiated:

        tan(9x)=sin(9x)cos(9x)\tan{\left(9 x \right)} = \frac{\sin{\left(9 x \right)}}{\cos{\left(9 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(9x)f{\left(x \right)} = \sin{\left(9 x \right)} and g(x)=cos(9x)g{\left(x \right)} = \cos{\left(9 x \right)}.

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

        1. Let u=9xu = 9 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 ddx9x\frac{d}{d x} 9 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: 99

          The result of the chain rule is:

          9cos(9x)9 \cos{\left(9 x \right)}

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

        1. Let u=9xu = 9 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 ddx9x\frac{d}{d x} 9 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: 99

          The result of the chain rule is:

          9sin(9x)- 9 \sin{\left(9 x \right)}

        Now plug in to the quotient rule:

        9sin2(9x)+9cos2(9x)cos2(9x)\frac{9 \sin^{2}{\left(9 x \right)} + 9 \cos^{2}{\left(9 x \right)}}{\cos^{2}{\left(9 x \right)}}

      The result of the chain rule is:

      9sin2(9x)+9cos2(9x)cos2(9x)tan2(9x)- \frac{9 \sin^{2}{\left(9 x \right)} + 9 \cos^{2}{\left(9 x \right)}}{\cos^{2}{\left(9 x \right)} \tan^{2}{\left(9 x \right)}}

    Method #2

    1. Rewrite the function to be differentiated:

      cot(9x)=cos(9x)sin(9x)\cot{\left(9 x \right)} = \frac{\cos{\left(9 x \right)}}{\sin{\left(9 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(9x)f{\left(x \right)} = \cos{\left(9 x \right)} and g(x)=sin(9x)g{\left(x \right)} = \sin{\left(9 x \right)}.

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

      1. Let u=9xu = 9 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 ddx9x\frac{d}{d x} 9 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: 99

        The result of the chain rule is:

        9sin(9x)- 9 \sin{\left(9 x \right)}

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

      1. Let u=9xu = 9 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 ddx9x\frac{d}{d x} 9 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: 99

        The result of the chain rule is:

        9cos(9x)9 \cos{\left(9 x \right)}

      Now plug in to the quotient rule:

      9sin2(9x)9cos2(9x)sin2(9x)\frac{- 9 \sin^{2}{\left(9 x \right)} - 9 \cos^{2}{\left(9 x \right)}}{\sin^{2}{\left(9 x \right)}}

  2. Now simplify:

    9sin2(9x)- \frac{9}{\sin^{2}{\left(9 x \right)}}


The answer is:

9sin2(9x)- \frac{9}{\sin^{2}{\left(9 x \right)}}

The graph
02468-8-6-4-2-1010-50005000
The first derivative [src]
          2     
-9 - 9*cot (9*x)
9cot2(9x)9- 9 \cot^{2}{\left(9 x \right)} - 9
The second derivative [src]
    /       2     \         
162*\1 + cot (9*x)/*cot(9*x)
162(cot2(9x)+1)cot(9x)162 \left(\cot^{2}{\left(9 x \right)} + 1\right) \cot{\left(9 x \right)}
The third derivative [src]
      /       2     \ /         2     \
-1458*\1 + cot (9*x)/*\1 + 3*cot (9*x)/
1458(cot2(9x)+1)(3cot2(9x)+1)- 1458 \left(\cot^{2}{\left(9 x \right)} + 1\right) \left(3 \cot^{2}{\left(9 x \right)} + 1\right)
The graph
Derivative of y=ctg9x