Derivative of a*e^(t*(-b))*sin(w*t+f0)

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

   1. Let $u = f_{0} + t w$.

   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 $\frac{\partial}{\partial w} \left(f_{0} + t w\right)$:

      1. Differentiate $f_{0} + t w$ 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: $w$ goes to $1$

            So, the result is: $t$

         2. The derivative of the constant $f_{0}$ is zero.

         The result is: $t$

      The result of the chain rule is:

      $$t \cos{\left(f_{0} + t w \right)}$$

   So, the result is: $a t e^{- b t} \cos{\left(f_{0} + t w \right)}$

2. Now simplify:

   $$a t e^{- b t} \cos{\left(f_{0} + t w \right)}$$

The answer is:

$$a t e^{- b t} \cos{\left(f_{0} + t w \right)}$$