Derivative of y=sin4x+cosx

1. Differentiate $\sin{\left(4 x \right)} + \cos{\left(x \right)}$ term by term:

   1. Let $u = 4 x$.

   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 $\frac{d}{d x} 4 x$:

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

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

         So, the result is: $4$

      The result of the chain rule is:

      $$4 \cos{\left(4 x \right)}$$

   4. The derivative of cosine is negative sine:

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

   The result is: $- \sin{\left(x \right)} + 4 \cos{\left(4 x \right)}$

The answer is:

$$- \sin{\left(x \right)} + 4 \cos{\left(4 x \right)}$$