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Derivative of (1/4)*sin(2*x+3)

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

You have entered [src]
sin(2*x + 3)
------------
     4      
sin(2x+3)4\frac{\sin{\left(2 x + 3 \right)}}{4}
sin(2*x + 3)/4
Detail solution
  1. The derivative of a constant times a function is the constant times the derivative of the function.

    1. Let u=2x+3u = 2 x + 3.

    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 ddx(2x+3)\frac{d}{d x} \left(2 x + 3\right):

      1. Differentiate 2x+32 x + 3 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 33 is zero.

        The result is: 22

      The result of the chain rule is:

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

    So, the result is: cos(2x+3)2\frac{\cos{\left(2 x + 3 \right)}}{2}

  2. Now simplify:

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


The answer is:

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

The graph
02468-8-6-4-2-10101-1
The first derivative [src]
cos(2*x + 3)
------------
     2      
cos(2x+3)2\frac{\cos{\left(2 x + 3 \right)}}{2}
The second derivative [src]
-sin(3 + 2*x)
sin(2x+3)- \sin{\left(2 x + 3 \right)}
The third derivative [src]
-2*cos(3 + 2*x)
2cos(2x+3)- 2 \cos{\left(2 x + 3 \right)}