Derivative of sin^3(3x-5)

1. Let $u = \sin{\left(3 x - 5 \right)}$.

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

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \sin{\left(3 x - 5 \right)}$:

   1. Let $u = 3 x - 5$.

   2. The derivative of sine is cosine:

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(3 x - 5\right)$:

      1. Differentiate $3 x - 5$ 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: $x$ goes to $1$

            So, the result is: $3$

         2. The derivative of the constant $-5$ is zero.

         The result is: $3$

      The result of the chain rule is:

      $$3 \cos{\left(3 x - 5 \right)}$$

   The result of the chain rule is:

   $$9 \sin^{2}{\left(3 x - 5 \right)} \cos{\left(3 x - 5 \right)}$$

4. Now simplify:

   $$9 \sin^{2}{\left(3 x - 5 \right)} \cos{\left(3 x - 5 \right)}$$

The answer is:

$$9 \sin^{2}{\left(3 x - 5 \right)} \cos{\left(3 x - 5 \right)}$$