Derivative of sin^4x+cos^4x

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

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

   2. Apply the power rule: $u^{4}$ goes to $4 u^{3}$

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

      1. The derivative of sine is cosine:

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

      The result of the chain rule is:

      $$4 \sin^{3}{\left(x \right)} \cos{\left(x \right)}$$

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

   5. Apply the power rule: $u^{4}$ goes to $4 u^{3}$

   6. 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:

      $$- 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)}$$

   The result is: $4 \sin^{3}{\left(x \right)} \cos{\left(x \right)} - 4 \sin{\left(x \right)} \cos^{3}{\left(x \right)}$

2. Now simplify:

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

The answer is:

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