Mister Exam

Derivative of sin2xcos²x

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   
sin(2*x)*cos (x)
sin(2x)cos2(x)\sin{\left(2 x \right)} \cos^{2}{\left(x \right)}
d /            2   \
--\sin(2*x)*cos (x)/
dx                  
ddxsin(2x)cos2(x)\frac{d}{d x} \sin{\left(2 x \right)} \cos^{2}{\left(x \right)}
Detail solution
  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)=sin(2x)f{\left(x \right)} = \sin{\left(2 x \right)}; to find ddxf(x)\frac{d}{d x} f{\left(x \right)}:

    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)}

    g(x)=cos2(x)g{\left(x \right)} = \cos^{2}{\left(x \right)}; to find ddxg(x)\frac{d}{d x} g{\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)}

    The result is: 2sin(x)sin(2x)cos(x)+2cos2(x)cos(2x)- 2 \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} + 2 \cos^{2}{\left(x \right)} \cos{\left(2 x \right)}

  2. Now simplify:

    2cos(x)cos(3x)2 \cos{\left(x \right)} \cos{\left(3 x \right)}


The answer is:

2cos(x)cos(3x)2 \cos{\left(x \right)} \cos{\left(3 x \right)}

The graph
02468-8-6-4-2-10105-5
The first derivative [src]
     2                                       
2*cos (x)*cos(2*x) - 2*cos(x)*sin(x)*sin(2*x)
2sin(x)sin(2x)cos(x)+2cos2(x)cos(2x)- 2 \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} + 2 \cos^{2}{\left(x \right)} \cos{\left(2 x \right)}
The second derivative [src]
  //   2         2   \                 2                                       \
2*\\sin (x) - cos (x)/*sin(2*x) - 2*cos (x)*sin(2*x) - 4*cos(x)*cos(2*x)*sin(x)/
2(4sin(x)cos(x)cos(2x)2sin(2x)cos2(x)+(sin2(x)cos2(x))sin(2x))2 \left(- 4 \sin{\left(x \right)} \cos{\left(x \right)} \cos{\left(2 x \right)} - 2 \sin{\left(2 x \right)} \cos^{2}{\left(x \right)} + \left(\sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}\right) \sin{\left(2 x \right)}\right)
The third derivative [src]
  /       2                 /   2         2   \                                    \
4*\- 2*cos (x)*cos(2*x) + 3*\sin (x) - cos (x)/*cos(2*x) + 8*cos(x)*sin(x)*sin(2*x)/
4(8sin(x)sin(2x)cos(x)2cos2(x)cos(2x)+3(sin2(x)cos2(x))cos(2x))4 \cdot \left(8 \sin{\left(x \right)} \sin{\left(2 x \right)} \cos{\left(x \right)} - 2 \cos^{2}{\left(x \right)} \cos{\left(2 x \right)} + 3 \left(\sin^{2}{\left(x \right)} - \cos^{2}{\left(x \right)}\right) \cos{\left(2 x \right)}\right)
The graph
Derivative of sin2xcos²x