Derivative of sin^3x*cos^3x

1. Apply the product rule:

   $$\frac{d}{d x} f{\left(x \right)} g{\left(x \right)} = f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}$$

   $f{\left(x \right)} = \sin^{3}{\left(x \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

   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:

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

   $g{\left(x \right)} = \cos^{3}{\left(x \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

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

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

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

      1. The derivative of cosine is negative sine:

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

      The result of the chain rule is:

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

   The result is: $- 3 \sin^{4}{\left(x \right)} \cos^{2}{\left(x \right)} + 3 \sin^{2}{\left(x \right)} \cos^{4}{\left(x \right)}$

2. Now simplify:

   $$\frac{3 \cos{\left(2 x \right)}}{16} - \frac{3 \cos{\left(6 x \right)}}{16}$$

The answer is:

$$\frac{3 \cos{\left(2 x \right)}}{16} - \frac{3 \cos{\left(6 x \right)}}{16}$$