Derivative of y=e^cosx+ln5

1. Differentiate $e^{\cos{\left(x \right)}} + \log{\left(5 \right)}$ term by term:

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

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

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

      1. The derivative of cosine is negative sine:

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

      The result of the chain rule is:

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

   4. The derivative of the constant $\log{\left(5 \right)}$ is zero.

   The result is: $- e^{\cos{\left(x \right)}} \sin{\left(x \right)}$

The answer is:

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