Derivative of y=xcos(2x^3+1)

1. Apply the product rule:

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

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

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

   $g{\left(x \right)} = \cos{\left(2 x^{3} + 1 \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = 2 x^{3} + 1$.

   2. The derivative of cosine is negative sine:

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

   3. Then, apply the chain rule. Multiply by $\frac{d}{d x} \left(2 x^{3} + 1\right)$:

      1. Differentiate $2 x^{3} + 1$ 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: $x^{3}$ goes to $3 x^{2}$

            So, the result is: $6 x^{2}$

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

         The result is: $6 x^{2}$

      The result of the chain rule is:

      $$- 6 x^{2} \sin{\left(2 x^{3} + 1 \right)}$$

   The result is: $- 6 x^{3} \sin{\left(2 x^{3} + 1 \right)} + \cos{\left(2 x^{3} + 1 \right)}$

2. Now simplify:

   $$- 6 x^{3} \sin{\left(2 x^{3} + 1 \right)} + \cos{\left(2 x^{3} + 1 \right)}$$

The answer is:

$$- 6 x^{3} \sin{\left(2 x^{3} + 1 \right)} + \cos{\left(2 x^{3} + 1 \right)}$$