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4sin2x-5ctgx
The derivative of the function/ 4sin2x-5ctgx

Derivative of 4sin2x-5ctgx

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4*sin(2*x) - 5*cot(x)
4sin⁡(2x)−5cot⁡(x)4 \sin{\left(2 x \right)} - 5 \cot{\left(x \right)} 
4*sin(2*x) - 5*cot(x)
Detail solution

  1. Differentiate 4sin⁡(2x)−5cot⁡(x)4 \sin{\left(2 x \right)} - 5 \cot{\left(x \right)} term by term:

    1. 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: 8cos⁡(2x)8 \cos{\left(2 x \right)}

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

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

          1. Rewrite the function to be differentiated:

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

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

            1. The derivative of sine is cosine:

              ddxsin⁡(x)=cos⁡(x)\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}

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

            1. The derivative of cosine is negative sine:

              ddxcos⁡(x)=−sin⁡(x)\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}

            Now plug in to the quotient rule:

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

          The result of the chain rule is:

          −sin⁡2(x)+cos⁡2(x)cos⁡2(x)tan⁡2(x)- \frac{\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}

        Method #2

        1. Rewrite the function to be differentiated:

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

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

          1. The derivative of cosine is negative sine:

            ddxcos⁡(x)=−sin⁡(x)\frac{d}{d x} \cos{\left(x \right)} = - \sin{\left(x \right)}

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

          1. The derivative of sine is cosine:

            ddxsin⁡(x)=cos⁡(x)\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}

          Now plug in to the quotient rule:

          −sin⁡2(x)−cos⁡2(x)sin⁡2(x)\frac{- \sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}}{\sin^{2}{\left(x \right)}}

      So, the result is: 5(sin⁡2(x)+cos⁡2(x))cos⁡2(x)tan⁡2(x)\frac{5 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}}

    The result is: 5(sin⁡2(x)+cos⁡2(x))cos⁡2(x)tan⁡2(x)+8cos⁡(2x)\frac{5 \left(\sin^{2}{\left(x \right)} + \cos^{2}{\left(x \right)}\right)}{\cos^{2}{\left(x \right)} \tan^{2}{\left(x \right)}} + 8 \cos{\left(2 x \right)}

  2. Now simplify:

    −16sin⁡2(x)+8+5sin⁡2(x)- 16 \sin^{2}{\left(x \right)} + 8 + \frac{5}{\sin^{2}{\left(x \right)}}

The answer is:

−16sin⁡2(x)+8+5sin⁡2(x)- 16 \sin^{2}{\left(x \right)} + 8 + \frac{5}{\sin^{2}{\left(x \right)}}

The graph
02468-8-6-4-2-1010-5000050000
Plot the graph f(x)
Plot the graph f'(x)
The first derivative [src] 
         2                
5 + 5*cot (x) + 8*cos(2*x)
8cos⁡(2x)+5cot⁡2(x)+58 \cos{\left(2 x \right)} + 5 \cot^{2}{\left(x \right)} + 5
Simplify
The second derivative [src] 
   /               /       2   \       \
-2*\8*sin(2*x) + 5*\1 + cot (x)/*cot(x)/
−2(5(cot⁡2(x)+1)cot⁡(x)+8sin⁡(2x))- 2 \left(5 \left(\cot^{2}{\left(x \right)} + 1\right) \cot{\left(x \right)} + 8 \sin{\left(2 x \right)}\right)
Simplify
The third derivative [src] 
  /                              2                           \
  |                 /       2   \          2    /       2   \|
2*\-16*cos(2*x) + 5*\1 + cot (x)/  + 10*cot (x)*\1 + cot (x)//
2(5(cot⁡2(x)+1)2+10(cot⁡2(x)+1)cot⁡2(x)−16cos⁡(2x))2 \left(5 \left(\cot^{2}{\left(x \right)} + 1\right)^{2} + 10 \left(\cot^{2}{\left(x \right)} + 1\right) \cot^{2}{\left(x \right)} - 16 \cos{\left(2 x \right)}\right)
Simplify
The graph
Derivative of 4sin2x-5ctgx
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