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Derivative of sin(n/6+2*x)

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

You have entered [src]
   /n      \
sin|- + 2*x|
   \6      /
sin(n6+2x)\sin{\left(\frac{n}{6} + 2 x \right)}
sin(n/6 + 2*x)
Detail solution
  1. Let u=n6+2xu = \frac{n}{6} + 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 x(n6+2x)\frac{\partial}{\partial x} \left(\frac{n}{6} + 2 x\right):

    1. Differentiate n6+2x\frac{n}{6} + 2 x term by term:

      1. The derivative of the constant n6\frac{n}{6} is zero.

      2. 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 is: 22

    The result of the chain rule is:

    2cos(n6+2x)2 \cos{\left(\frac{n}{6} + 2 x \right)}

  4. Now simplify:

    2cos(n6+2x)2 \cos{\left(\frac{n}{6} + 2 x \right)}


The answer is:

2cos(n6+2x)2 \cos{\left(\frac{n}{6} + 2 x \right)}

The first derivative [src]
     /n      \
2*cos|- + 2*x|
     \6      /
2cos(n6+2x)2 \cos{\left(\frac{n}{6} + 2 x \right)}
The second derivative [src]
      /      n\
-4*sin|2*x + -|
      \      6/
4sin(n6+2x)- 4 \sin{\left(\frac{n}{6} + 2 x \right)}
The third derivative [src]
      /      n\
-8*cos|2*x + -|
      \      6/
8cos(n6+2x)- 8 \cos{\left(\frac{n}{6} + 2 x \right)}