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cos^2(2x+1)

Derivative of cos^2(2x+1)

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

The graph:

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Piecewise:

The solution

You have entered [src]
   2         
cos (2*x + 1)
cos2(2x+1)\cos^{2}{\left(2 x + 1 \right)}
d /   2         \
--\cos (2*x + 1)/
dx               
ddxcos2(2x+1)\frac{d}{d x} \cos^{2}{\left(2 x + 1 \right)}
Detail solution
  1. Let u=cos(2x+1)u = \cos{\left(2 x + 1 \right)}.

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

  3. Then, apply the chain rule. Multiply by ddxcos(2x+1)\frac{d}{d x} \cos{\left(2 x + 1 \right)}:

    1. Let u=2x+1u = 2 x + 1.

    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(2x+1)\frac{d}{d x} \left(2 x + 1\right):

      1. Differentiate 2x+12 x + 1 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: xx goes to 11

          So, the result is: 22

        2. The derivative of the constant 11 is zero.

        The result is: 22

      The result of the chain rule is:

      2sin(2x+1)- 2 \sin{\left(2 x + 1 \right)}

    The result of the chain rule is:

    4sin(2x+1)cos(2x+1)- 4 \sin{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}

  4. Now simplify:

    2sin(4x+2)- 2 \sin{\left(4 x + 2 \right)}


The answer is:

2sin(4x+2)- 2 \sin{\left(4 x + 2 \right)}

The graph
02468-8-6-4-2-10105-5
The first derivative [src]
-4*cos(2*x + 1)*sin(2*x + 1)
4sin(2x+1)cos(2x+1)- 4 \sin{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}
The second derivative [src]
  /   2               2         \
8*\sin (1 + 2*x) - cos (1 + 2*x)/
8(sin2(2x+1)cos2(2x+1))8 \left(\sin^{2}{\left(2 x + 1 \right)} - \cos^{2}{\left(2 x + 1 \right)}\right)
The third derivative [src]
64*cos(1 + 2*x)*sin(1 + 2*x)
64sin(2x+1)cos(2x+1)64 \sin{\left(2 x + 1 \right)} \cos{\left(2 x + 1 \right)}
The graph
Derivative of cos^2(2x+1)