Derivative of e^(-sin(t))

1. Let $u = - \sin{\left(t \right)}$.

2. The derivative of $e^{u}$ is itself.

3. Then, apply the chain rule. Multiply by $\frac{d}{d t} \left(- \sin{\left(t \right)}\right)$:

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

      1. The derivative of sine is cosine:

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

      So, the result is: $- \cos{\left(t \right)}$

   The result of the chain rule is:

   $$- e^{- \sin{\left(t \right)}} \cos{\left(t \right)}$$

The answer is:

$$- e^{- \sin{\left(t \right)}} \cos{\left(t \right)}$$