Derivative of Сsin(at+b)

1. The derivative of a constant times a function is the constant times the derivative of the function.

   1. Let $u = a t + b$.

   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 $\frac{\partial}{\partial t} \left(a t + b\right)$:

      1. Differentiate $a t + b$ 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: $t$ goes to $1$

            So, the result is: $a$

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

         The result is: $a$

      The result of the chain rule is:

      $$a \cos{\left(a t + b \right)}$$

   So, the result is: $a c \cos{\left(a t + b \right)}$

2. Now simplify:

   $$a c \cos{\left(a t + b \right)}$$

The answer is:

$$a c \cos{\left(a t + b \right)}$$