We are given the function $y = \sin x \cos x$, and we need to find the derivative $\frac{dy}{dx}$ at $x = \frac{\pi}{3}$.

Step 1: Differentiate $y = \sin x \cos x$

We can apply the product rule for differentiation here. The product rule states that if $y = u(x) v(x)$, then:

$\frac{dy}{dx} = u'(x) v(x) + u(x) v'(x)$

In this case:

• $u(x) = \sin x$, so $u'(x) = \cos x$
• $v(x) = \cos x$, so $v'(x) = -\sin x$

Now, applying the product rule:

$\frac{dy}{dx} = \cos x \cdot \cos x + \sin x \cdot (-\sin x)$ $\frac{dy}{dx} = \cos^2 x - \sin^2 x$

Step 2: Substitute $x = \frac{\pi}{3}$

We now substitute $x = \frac{\pi}{3}$ into the derivative.

• $\cos \frac{\pi}{3} = \frac{1}{2}$
• $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$

Substituting these values into the expression for $\frac{dy}{dx}$:

$\frac{dy}{dx} = \left( \frac{1}{2} \right)^2 - \left( \frac{\sqrt{3}}{2} \right)^2$ $\frac{dy}{dx} = \frac{1}{4} - \frac{3}{4}$ $\frac{dy}{dx} = \frac{-2}{4} = -\frac{1}{2}$

Conclusion:

At $x = \frac{\pi}{3}$, the derivative $\frac{dy}{dx} = -\frac{1}{2}$.

Thus, the correct answer is: $\boxed{-\frac{1}{2}} \quad \text{(option a)}$

Would you like further details on any part of the solution or have any questions?

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Here are some related questions you might find useful:

1. What is the derivative of $\sin^2 x$?
2. How does the product rule differ from the chain rule in differentiation?
3. Can you prove the identity $\cos^2 x - \sin^2 x = \cos 2x$?
4. What is the second derivative of $y = \sin x \cos x$?
5. How would you integrate $\sin x \cos x$?

Tip: For trigonometric products, it’s often useful to express them using trigonometric identities to simplify differentiation or integration.