Derivative of (4x-9)cos(x/5)

1. Apply the product rule:

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

   $f{\left(x \right)} = 4 x - 9$; to find $\frac{d}{d x} f{\left(x \right)}$:

   1. Differentiate $4 x - 9$ 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: $4$

      2. The derivative of the constant $-9$ is zero.

      The result is: $4$

   $g{\left(x \right)} = \cos{\left(\frac{x}{5} \right)}$; to find $\frac{d}{d x} g{\left(x \right)}$:

   1. Let $u = \frac{x}{5}$.

   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 $\frac{d}{d x} \frac{x}{5}$:

      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: $\frac{1}{5}$

      The result of the chain rule is:

      $$- \frac{\sin{\left(\frac{x}{5} \right)}}{5}$$

   The result is: $- \frac{\left(4 x - 9\right) \sin{\left(\frac{x}{5} \right)}}{5} + 4 \cos{\left(\frac{x}{5} \right)}$

2. Now simplify:

   $$\frac{\left(9 - 4 x\right) \sin{\left(\frac{x}{5} \right)}}{5} + 4 \cos{\left(\frac{x}{5} \right)}$$

The answer is:

$$\frac{\left(9 - 4 x\right) \sin{\left(\frac{x}{5} \right)}}{5} + 4 \cos{\left(\frac{x}{5} \right)}$$