Derivative of exp(3x)(x+1)

1. Apply the product rule:

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

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

   1. Let $u = 3 x$.

   2. The derivative of $e^{u}$ is itself.

   3. 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:

      $$3 e^{3 x}$$

   $g{\left(x \right)} = x + 1$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Differentiate $x + 1$ term by term:

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

      2. The derivative of the constant $1$ is zero.

      The result is: $1$

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

2. Now simplify:

   $$\left(3 x + 4\right) e^{3 x}$$

The answer is:

$$\left(3 x + 4\right) e^{3 x}$$