Derivative of sin(3*x-1)*e^x

1. Apply the product rule:

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

   $f{\left(x \right)} = \sin{\left(3 x - 1 \right)}$; to find $\frac{d}{d x} f{\left(x \right)}$:

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

   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 $\frac{d}{d x} \left(3 x - 1\right)$:

      1. Differentiate $3 x - 1$ 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 $-1$ is zero.

         The result is: $3$

      The result of the chain rule is:

      $$3 \cos{\left(3 x - 1 \right)}$$

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

   1. The derivative of $e^{x}$ is itself.

   The result is: $e^{x} \sin{\left(3 x - 1 \right)} + 3 e^{x} \cos{\left(3 x - 1 \right)}$

2. Now simplify:

   $$\left(\sin{\left(3 x - 1 \right)} + 3 \cos{\left(3 x - 1 \right)}\right) e^{x}$$

The answer is:

$$\left(\sin{\left(3 x - 1 \right)} + 3 \cos{\left(3 x - 1 \right)}\right) e^{x}$$