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y=sin(ctg(9x^2+5))

Derivative of y=sin(ctg(9x^2+5))

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sin\cot\9*x  + 5//
sin(cot(9x2+5))\sin{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}
sin(cot(9*x^2 + 5))
Detail solution
  1. Let u=cot(9x2+5)u = \cot{\left(9 x^{2} + 5 \right)}.

  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 ddxcot(9x2+5)\frac{d}{d x} \cot{\left(9 x^{2} + 5 \right)}:

    1. There are multiple ways to do this derivative.

      Method #1

      1. Rewrite the function to be differentiated:

        cot(9x2+5)=1tan(9x2+5)\cot{\left(9 x^{2} + 5 \right)} = \frac{1}{\tan{\left(9 x^{2} + 5 \right)}}

      2. Let u=tan(9x2+5)u = \tan{\left(9 x^{2} + 5 \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(9x2+5)\frac{d}{d x} \tan{\left(9 x^{2} + 5 \right)}:

        1. Rewrite the function to be differentiated:

          tan(9x2+5)=sin(9x2+5)cos(9x2+5)\tan{\left(9 x^{2} + 5 \right)} = \frac{\sin{\left(9 x^{2} + 5 \right)}}{\cos{\left(9 x^{2} + 5 \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(9x2+5)f{\left(x \right)} = \sin{\left(9 x^{2} + 5 \right)} and g(x)=cos(9x2+5)g{\left(x \right)} = \cos{\left(9 x^{2} + 5 \right)}.

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

          1. Let u=9x2+5u = 9 x^{2} + 5.

          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 ddx(9x2+5)\frac{d}{d x} \left(9 x^{2} + 5\right):

            1. Differentiate 9x2+59 x^{2} + 5 term by term:

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

                1. Apply the power rule: x2x^{2} goes to 2x2 x

                So, the result is: 18x18 x

              2. The derivative of the constant 55 is zero.

              The result is: 18x18 x

            The result of the chain rule is:

            18xcos(9x2+5)18 x \cos{\left(9 x^{2} + 5 \right)}

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

          1. Let u=9x2+5u = 9 x^{2} + 5.

          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 ddx(9x2+5)\frac{d}{d x} \left(9 x^{2} + 5\right):

            1. Differentiate 9x2+59 x^{2} + 5 term by term:

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

                1. Apply the power rule: x2x^{2} goes to 2x2 x

                So, the result is: 18x18 x

              2. The derivative of the constant 55 is zero.

              The result is: 18x18 x

            The result of the chain rule is:

            18xsin(9x2+5)- 18 x \sin{\left(9 x^{2} + 5 \right)}

          Now plug in to the quotient rule:

          18xsin2(9x2+5)+18xcos2(9x2+5)cos2(9x2+5)\frac{18 x \sin^{2}{\left(9 x^{2} + 5 \right)} + 18 x \cos^{2}{\left(9 x^{2} + 5 \right)}}{\cos^{2}{\left(9 x^{2} + 5 \right)}}

        The result of the chain rule is:

        18xsin2(9x2+5)+18xcos2(9x2+5)cos2(9x2+5)tan2(9x2+5)- \frac{18 x \sin^{2}{\left(9 x^{2} + 5 \right)} + 18 x \cos^{2}{\left(9 x^{2} + 5 \right)}}{\cos^{2}{\left(9 x^{2} + 5 \right)} \tan^{2}{\left(9 x^{2} + 5 \right)}}

      Method #2

      1. Rewrite the function to be differentiated:

        cot(9x2+5)=cos(9x2+5)sin(9x2+5)\cot{\left(9 x^{2} + 5 \right)} = \frac{\cos{\left(9 x^{2} + 5 \right)}}{\sin{\left(9 x^{2} + 5 \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(9x2+5)f{\left(x \right)} = \cos{\left(9 x^{2} + 5 \right)} and g(x)=sin(9x2+5)g{\left(x \right)} = \sin{\left(9 x^{2} + 5 \right)}.

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

        1. Let u=9x2+5u = 9 x^{2} + 5.

        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 ddx(9x2+5)\frac{d}{d x} \left(9 x^{2} + 5\right):

          1. Differentiate 9x2+59 x^{2} + 5 term by term:

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

              1. Apply the power rule: x2x^{2} goes to 2x2 x

              So, the result is: 18x18 x

            2. The derivative of the constant 55 is zero.

            The result is: 18x18 x

          The result of the chain rule is:

          18xsin(9x2+5)- 18 x \sin{\left(9 x^{2} + 5 \right)}

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

        1. Let u=9x2+5u = 9 x^{2} + 5.

        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 ddx(9x2+5)\frac{d}{d x} \left(9 x^{2} + 5\right):

          1. Differentiate 9x2+59 x^{2} + 5 term by term:

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

              1. Apply the power rule: x2x^{2} goes to 2x2 x

              So, the result is: 18x18 x

            2. The derivative of the constant 55 is zero.

            The result is: 18x18 x

          The result of the chain rule is:

          18xcos(9x2+5)18 x \cos{\left(9 x^{2} + 5 \right)}

        Now plug in to the quotient rule:

        18xsin2(9x2+5)18xcos2(9x2+5)sin2(9x2+5)\frac{- 18 x \sin^{2}{\left(9 x^{2} + 5 \right)} - 18 x \cos^{2}{\left(9 x^{2} + 5 \right)}}{\sin^{2}{\left(9 x^{2} + 5 \right)}}

    The result of the chain rule is:

    (18xsin2(9x2+5)+18xcos2(9x2+5))cos(cot(9x2+5))cos2(9x2+5)tan2(9x2+5)- \frac{\left(18 x \sin^{2}{\left(9 x^{2} + 5 \right)} + 18 x \cos^{2}{\left(9 x^{2} + 5 \right)}\right) \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}}{\cos^{2}{\left(9 x^{2} + 5 \right)} \tan^{2}{\left(9 x^{2} + 5 \right)}}

  4. Now simplify:

    18xcos(cot(9x2+5))cos2(9x2+5)tan2(9x2+5)- \frac{18 x \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}}{\cos^{2}{\left(9 x^{2} + 5 \right)} \tan^{2}{\left(9 x^{2} + 5 \right)}}


The answer is:

18xcos(cot(9x2+5))cos2(9x2+5)tan2(9x2+5)- \frac{18 x \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}}{\cos^{2}{\left(9 x^{2} + 5 \right)} \tan^{2}{\left(9 x^{2} + 5 \right)}}

The graph
02468-8-6-4-2-1010-500000500000
The first derivative [src]
     /        2/   2    \\    /   /   2    \\
18*x*\-1 - cot \9*x  + 5//*cos\cot\9*x  + 5//
18x(cot2(9x2+5)1)cos(cot(9x2+5))18 x \left(- \cot^{2}{\left(9 x^{2} + 5 \right)} - 1\right) \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}
The second derivative [src]
   /       2/       2\\ /     /   /       2\\       2 /       2/       2\\    /   /       2\\       2    /   /       2\\    /       2\\
18*\1 + cot \5 + 9*x //*\- cos\cot\5 + 9*x // - 18*x *\1 + cot \5 + 9*x //*sin\cot\5 + 9*x // + 36*x *cos\cot\5 + 9*x //*cot\5 + 9*x //
18(cot2(9x2+5)+1)(18x2(cot2(9x2+5)+1)sin(cot(9x2+5))+36x2cos(cot(9x2+5))cot(9x2+5)cos(cot(9x2+5)))18 \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \left(- 18 x^{2} \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \sin{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} + 36 x^{2} \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} \cot{\left(9 x^{2} + 5 \right)} - \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)}\right)
The third derivative [src]
                           /                                                                                                                                                                                                    2                                                                                 \
      /       2/       2\\ |  /       2/       2\\    /   /       2\\        /   /       2\\    /       2\       2    2/       2\    /   /       2\\       2 /       2/       2\\    /   /       2\\      2 /       2/       2\\     /   /       2\\       2 /       2/       2\\    /       2\    /   /       2\\|
972*x*\1 + cot \5 + 9*x //*\- \1 + cot \5 + 9*x //*sin\cot\5 + 9*x // + 2*cos\cot\5 + 9*x //*cot\5 + 9*x / - 24*x *cot \5 + 9*x /*cos\cot\5 + 9*x // - 12*x *\1 + cot \5 + 9*x //*cos\cot\5 + 9*x // + 6*x *\1 + cot \5 + 9*x // *cos\cot\5 + 9*x // + 36*x *\1 + cot \5 + 9*x //*cot\5 + 9*x /*sin\cot\5 + 9*x ///
972x(cot2(9x2+5)+1)(6x2(cot2(9x2+5)+1)2cos(cot(9x2+5))+36x2(cot2(9x2+5)+1)sin(cot(9x2+5))cot(9x2+5)12x2(cot2(9x2+5)+1)cos(cot(9x2+5))24x2cos(cot(9x2+5))cot2(9x2+5)(cot2(9x2+5)+1)sin(cot(9x2+5))+2cos(cot(9x2+5))cot(9x2+5))972 x \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \left(6 x^{2} \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right)^{2} \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} + 36 x^{2} \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \sin{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} \cot{\left(9 x^{2} + 5 \right)} - 12 x^{2} \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} - 24 x^{2} \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} \cot^{2}{\left(9 x^{2} + 5 \right)} - \left(\cot^{2}{\left(9 x^{2} + 5 \right)} + 1\right) \sin{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} + 2 \cos{\left(\cot{\left(9 x^{2} + 5 \right)} \right)} \cot{\left(9 x^{2} + 5 \right)}\right)
The graph
Derivative of y=sin(ctg(9x^2+5))