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Derivative of 2*sin(t)*cos(t)+1

Function f() - derivative -N order at the point
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2*sin(t)*cos(t) + 1
2sin(t)cos(t)+12 \sin{\left(t \right)} \cos{\left(t \right)} + 1
(2*sin(t))*cos(t) + 1
Detail solution
  1. Differentiate 2sin(t)cos(t)+12 \sin{\left(t \right)} \cos{\left(t \right)} + 1 term by term:

    1. Apply the product rule:

      ddtf(t)g(t)=f(t)ddtg(t)+g(t)ddtf(t)\frac{d}{d t} f{\left(t \right)} g{\left(t \right)} = f{\left(t \right)} \frac{d}{d t} g{\left(t \right)} + g{\left(t \right)} \frac{d}{d t} f{\left(t \right)}

      f(t)=2sin(t)f{\left(t \right)} = 2 \sin{\left(t \right)}; to find ddtf(t)\frac{d}{d t} f{\left(t \right)}:

      1. The derivative of a constant times a function is the constant times the derivative of the function.

        1. The derivative of sine is cosine:

          ddtsin(t)=cos(t)\frac{d}{d t} \sin{\left(t \right)} = \cos{\left(t \right)}

        So, the result is: 2cos(t)2 \cos{\left(t \right)}

      g(t)=cos(t)g{\left(t \right)} = \cos{\left(t \right)}; to find ddtg(t)\frac{d}{d t} g{\left(t \right)}:

      1. The derivative of cosine is negative sine:

        ddtcos(t)=sin(t)\frac{d}{d t} \cos{\left(t \right)} = - \sin{\left(t \right)}

      The result is: 2sin2(t)+2cos2(t)- 2 \sin^{2}{\left(t \right)} + 2 \cos^{2}{\left(t \right)}

    2. The derivative of the constant 11 is zero.

    The result is: 2sin2(t)+2cos2(t)- 2 \sin^{2}{\left(t \right)} + 2 \cos^{2}{\left(t \right)}

  2. Now simplify:

    2cos(2t)2 \cos{\left(2 t \right)}


The answer is:

2cos(2t)2 \cos{\left(2 t \right)}

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