To solve this problem, we need to find the derivative of the given function $f(x)$.

The function is: $f(x) = \sin^2(x) - 2\sin(x)$

Step 1: Derivative of $\sin^2(x)$

Using the chain rule, the derivative of $\sin^2(x)$ is: $\frac{d}{dx}[\sin^2(x)] = 2\sin(x) \cdot \cos(x) = \sin(2x)$

Step 2: Derivative of $-2\sin(x)$

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

Step 3: Combine the results

Now, the derivative of $f(x)$ is: $f'(x) = \sin(2x) - 2\cos(x)$

Final Answer:

From the options provided, the correct answer is: $\boxed{\text{E. } \sin(2x) - 2\cos(x)}$

Would you like further explanation or related derivations?

Related Questions:

1. What is the derivative of $\cos^2(x)$?
2. Can you explain the chain rule in detail for trigonometric functions?
3. How does the product rule apply to trigonometric identities like $\sin^2(x)$?
4. What is the second derivative of $f(x) = \sin^2(x) - 2\sin(x)$?
5. How do you verify that $\sin(2x) = 2\sin(x)\cos(x)$ is a valid trigonometric identity?

Tip:

Remember that the derivative of a composite function $u(v(x))$ requires the chain rule: first, take the derivative of the outer function, then multiply by the derivative of the inner function.