Derivative of sin(3*x)/(x+1)

1. Apply the quotient rule, which is:

   $$\frac{d}{d x} \frac{f{\left(x \right)}}{g{\left(x \right)}} = \frac{- f{\left(x \right)} \frac{d}{d x} g{\left(x \right)} + g{\left(x \right)} \frac{d}{d x} f{\left(x \right)}}{g^{2}{\left(x \right)}}$$

   $f{\left(x \right)} = \sin{\left(3 x \right)}$ and $g{\left(x \right)} = x + 1$.

   To find $\frac{d}{d x} f{\left(x \right)}$:

   1. Let $u = 3 x$.

   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 $\frac{d}{d x} 3 x$:

      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:

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

   To find $\frac{d}{d x} g{\left(x \right)}$:

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

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

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

      The result is: $1$

   Now plug in to the quotient rule:

   $\frac{3 \left(x + 1\right) \cos{\left(3 x \right)} - \sin{\left(3 x \right)}}{\left(x + 1\right)^{2}}$

2. Now simplify:

   $$\frac{\left(3 x + 3\right) \cos{\left(3 x \right)} - \sin{\left(3 x \right)}}{\left(x + 1\right)^{2}}$$

The answer is:

$$\frac{\left(3 x + 3\right) \cos{\left(3 x \right)} - \sin{\left(3 x \right)}}{\left(x + 1\right)^{2}}$$