To find the derivative of $f(x) = \sin(x) \cos(x)$, we use the product rule:

Product Rule:

If $f(x) = u(x) \cdot v(x)$, then
$f'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x).$

Here, $u(x) = \sin(x)$ and $v(x) = \cos(x)$.

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Step 1: Differentiate $u(x) = \sin(x)$

$u'(x) = \cos(x).$

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Step 2: Differentiate $v(x) = \cos(x)$

$v'(x) = -\sin(x).$

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Step 3: Apply the product rule

$f'(x) = (\cos(x)) \cdot (\cos(x)) + (\sin(x)) \cdot (-\sin(x)).$

Simplify: $f'(x) = \cos^2(x) - \sin^2(x).$

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Final Answer:

$f'(x) = \cos^2(x) - \sin^2(x).$

This result can also be rewritten using the double-angle identity:
$\cos(2x) = \cos^2(x) - \sin^2(x).$
Thus: $f'(x) = \cos(2x).$

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Would you like a step-by-step breakdown of how the double-angle identity is derived?

Related Questions:

1. How do you derive the product rule?
2. Why is $\cos(2x) = \cos^2(x) - \sin^2(x)$?
3. How do you use the chain rule for trigonometric functions?
4. What are the steps to differentiate higher powers of trigonometric products, such as $(\sin(x)\cos(x))^2$?
5. How does the derivative of $\sin(x)\cos(x)$ relate to its integral?

Tip:

Remember that trigonometric identities often simplify derivatives, so it's useful to recognize them quickly!