Derivative of e^(4cos(x-1))

1. Let $u = 4 \cos{\left(x - 1 \right)}$.

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

3. Then, apply the chain rule. Multiply by $\frac{d}{d x} 4 \cos{\left(x - 1 \right)}$:

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

      1. Let $u = x - 1$.

      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 $\frac{d}{d x} \left(x - 1\right)$:

         1. Differentiate $x - 1$ term by term:

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

            2. The derivative of the constant $\left(-1\right) 1$ is zero.

            The result is: $1$

         The result of the chain rule is:

         $$- \sin{\left(x - 1 \right)}$$

      So, the result is: $- 4 \sin{\left(x - 1 \right)}$

   The result of the chain rule is:

   $$- 4 e^{4 \cos{\left(x - 1 \right)}} \sin{\left(x - 1 \right)}$$

4. Now simplify:

   $$- 4 e^{4 \cos{\left(x - 1 \right)}} \sin{\left(x - 1 \right)}$$

The answer is:

$$- 4 e^{4 \cos{\left(x - 1 \right)}} \sin{\left(x - 1 \right)}$$