Mister Exam

Derivative of y=ctg(5x)

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

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The solution

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cot(5*x)
cot(5x)\cot{\left(5 x \right)}
d           
--(cot(5*x))
dx          
ddxcot(5x)\frac{d}{d x} \cot{\left(5 x \right)}
Detail solution
  1. There are multiple ways to do this derivative.

    Method #1

    1. Rewrite the function to be differentiated:

      cot(5x)=1tan(5x)\cot{\left(5 x \right)} = \frac{1}{\tan{\left(5 x \right)}}

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

      1. Rewrite the function to be differentiated:

        tan(5x)=sin(5x)cos(5x)\tan{\left(5 x \right)} = \frac{\sin{\left(5 x \right)}}{\cos{\left(5 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(5x)f{\left(x \right)} = \sin{\left(5 x \right)} and g(x)=cos(5x)g{\left(x \right)} = \cos{\left(5 x \right)}.

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

        1. Let u=5xu = 5 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 ddx5x\frac{d}{d x} 5 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: 55

          The result of the chain rule is:

          5cos(5x)5 \cos{\left(5 x \right)}

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

        1. Let u=5xu = 5 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 ddx5x\frac{d}{d x} 5 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: 55

          The result of the chain rule is:

          5sin(5x)- 5 \sin{\left(5 x \right)}

        Now plug in to the quotient rule:

        5sin2(5x)+5cos2(5x)cos2(5x)\frac{5 \sin^{2}{\left(5 x \right)} + 5 \cos^{2}{\left(5 x \right)}}{\cos^{2}{\left(5 x \right)}}

      The result of the chain rule is:

      5sin2(5x)+5cos2(5x)cos2(5x)tan2(5x)- \frac{5 \sin^{2}{\left(5 x \right)} + 5 \cos^{2}{\left(5 x \right)}}{\cos^{2}{\left(5 x \right)} \tan^{2}{\left(5 x \right)}}

    Method #2

    1. Rewrite the function to be differentiated:

      cot(5x)=cos(5x)sin(5x)\cot{\left(5 x \right)} = \frac{\cos{\left(5 x \right)}}{\sin{\left(5 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(5x)f{\left(x \right)} = \cos{\left(5 x \right)} and g(x)=sin(5x)g{\left(x \right)} = \sin{\left(5 x \right)}.

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

      1. Let u=5xu = 5 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 ddx5x\frac{d}{d x} 5 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: 55

        The result of the chain rule is:

        5sin(5x)- 5 \sin{\left(5 x \right)}

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

      1. Let u=5xu = 5 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 ddx5x\frac{d}{d x} 5 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: 55

        The result of the chain rule is:

        5cos(5x)5 \cos{\left(5 x \right)}

      Now plug in to the quotient rule:

      5sin2(5x)5cos2(5x)sin2(5x)\frac{- 5 \sin^{2}{\left(5 x \right)} - 5 \cos^{2}{\left(5 x \right)}}{\sin^{2}{\left(5 x \right)}}

  2. Now simplify:

    5sin2(5x)- \frac{5}{\sin^{2}{\left(5 x \right)}}


The answer is:

5sin2(5x)- \frac{5}{\sin^{2}{\left(5 x \right)}}

The graph
02468-8-6-4-2-1010-50005000
The first derivative [src]
          2     
-5 - 5*cot (5*x)
5cot2(5x)5- 5 \cot^{2}{\left(5 x \right)} - 5
The second derivative [src]
   /       2     \         
50*\1 + cot (5*x)/*cot(5*x)
50(cot2(5x)+1)cot(5x)50 \left(\cot^{2}{\left(5 x \right)} + 1\right) \cot{\left(5 x \right)}
The third derivative [src]
     /       2     \ /         2     \
-250*\1 + cot (5*x)/*\1 + 3*cot (5*x)/
250(cot2(5x)+1)(3cot2(5x)+1)- 250 \left(\cot^{2}{\left(5 x \right)} + 1\right) \left(3 \cot^{2}{\left(5 x \right)} + 1\right)
The graph
Derivative of y=ctg(5x)