Mister Exam

Derivative of 2sin^2xcos^2x

Function f() - derivative -N order at the point
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The graph:

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The solution

You have entered [src]
     2       2   
2*sin (x)*cos (x)
2sin2(x)cos2(x)2 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}
d /     2       2   \
--\2*sin (x)*cos (x)/
dx                   
ddx2sin2(x)cos2(x)\frac{d}{d x} 2 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)}
Detail solution
  1. The derivative of a constant times a function is the constant times the derivative of the function.

    1. Apply the product rule:

      ddxf(x)g(x)=f(x)ddxg(x)+g(x)ddxf(x)\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}

      f(x)=cos2(x)f{\left(x \right)} = \cos^{2}{\left(x \right)}; to find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

      1. Let u=cos(x)u = \cos{\left(x \right)}.

      2. Apply the power rule: u2u^{2} goes to 2u2 u

      3. Then, apply the chain rule. Multiply by ddxcos(x)\frac{d}{d x} \cos{\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)}

        The result of the chain rule is:

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

      g(x)=sin2(x)g{\left(x \right)} = \sin^{2}{\left(x \right)}; to find ddxg(x)\frac{d}{d x} g{\left(x \right)}:

      1. Let u=sin(x)u = \sin{\left(x \right)}.

      2. Apply the power rule: u2u^{2} goes to 2u2 u

      3. Then, apply the chain rule. Multiply by ddxsin(x)\frac{d}{d x} \sin{\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)}

        The result of the chain rule is:

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

      The result is: 2sin3(x)cos(x)+2sin(x)cos3(x)- 2 \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 2 \sin{\left(x \right)} \cos^{3}{\left(x \right)}

    So, the result is: 4sin3(x)cos(x)+4sin(x)cos3(x)- 4 \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)}

  2. Now simplify:

    sin(4x)\sin{\left(4 x \right)}


The answer is:

sin(4x)\sin{\left(4 x \right)}

The graph
02468-8-6-4-2-10102-2
The first derivative [src]
       3                  3          
- 4*sin (x)*cos(x) + 4*cos (x)*sin(x)
4sin3(x)cos(x)+4sin(x)cos3(x)- 4 \sin^{3}{\left(x \right)} \cos{\left(x \right)} + 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)}
The second derivative [src]
  /   2    /   2         2   \      2    /   2         2   \        2       2   \
4*\sin (x)*\sin (x) - cos (x)/ - cos (x)*\sin (x) - cos (x)/ - 4*cos (x)*sin (x)/
4(4sin2(x)cos2(x)+(sin2(x)cos2(x))sin2(x)(sin2(x)cos2(x))cos2(x))4 \left(- 4 \sin^{2}{\left(x \right)} \cos^{2}{\left(x \right)} + \left(\sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}\right) \sin^{2}{\left(x \right)} - \left(\sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}\right) \cos^{2}{\left(x \right)}\right)
The third derivative [src]
   /       2           2   \              
16*\- 4*cos (x) + 4*sin (x)/*cos(x)*sin(x)
16(4sin2(x)4cos2(x))sin(x)cos(x)16 \cdot \left(4 \sin^{2}{\left(x \right)} - 4 \cos^{2}{\left(x \right)}\right) \sin{\left(x \right)} \cos{\left(x \right)}
The graph
Derivative of 2sin^2xcos^2x