To compute the derivative of $2\sin(x) + \cos(2x)$, we differentiate each term separately.

1. The derivative of $2\sin(x)$: $\frac{d}{dx}[2\sin(x)] = 2\cos(x)$

2. The derivative of $\cos(2x)$ (using the chain rule): [ \frac{d}{dx}[\cos(2x)] = -\sin(2x) \cdot \frac{d}{dx}[2x] = -2\sin(2x) ]

Now combine the results: $\frac{d}{dx}[2\sin(x) + \cos(2x)] = 2\cos(x) - 2\sin(2x)$

Final Answer:

$\boxed{2\cos(x) - 2\sin(2x)}$

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Would you like further details or assistance? Below are some related questions for practice:

1. What is the derivative of $\cos(3x) + 5\sin(x)$?
2. How do you simplify trigonometric expressions like $\sin(2x)$ using identities?
3. What is the second derivative of $2\sin(x) + \cos(2x)$?
4. How do you compute the derivative of $\tan(x) + \cot(x)$?
5. Why does the chain rule apply to $\cos(2x)$, and how do you use it?

Tip: When differentiating trigonometric functions involving multiples of $x$, always apply the chain rule to handle the inner function!