Derivative of y=sin^3x+sin3x

1. Differentiate $\sin^{3}{\left(x \right)} + \sin{\left(3 x \right)}$ term by term:

   1. Let $u = \sin{\left(x \right)}$.

   2. Apply the power rule: $u^{3}$ goes to $3 u^{2}$

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sin{\left(x \right)}$:

      1. The derivative of sine is cosine:

         $$\frac{d}{d x} \sin{\left(x \right)} = \cos{\left(x \right)}$$

      The result of the chain rule is:

      $$3 \sin^{2}{\left(x \right)} \cos{\left(x \right)}$$

   4. Let $u = 3 x$.

   5. The derivative of sine is cosine:

      $$\frac{d}{d u} \sin{\left(u \right)} = \cos{\left(u \right)}$$

   6. Then, apply the chain rule. Multiply by $\frac{d}{d x} 3 x$:

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

         1. Apply the power rule: $x$ goes to $1$

         So, the result is: $3$

      The result of the chain rule is:

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

   The result is: $3 \sin^{2}{\left(x \right)} \cos{\left(x \right)} + 3 \cos{\left(3 x \right)}$

2. Now simplify:

   $$\frac{3 \cos{\left(x \right)}}{4} + \frac{9 \cos{\left(3 x \right)}}{4}$$

The answer is:

$$\frac{3 \cos{\left(x \right)}}{4} + \frac{9 \cos{\left(3 x \right)}}{4}$$