How do you use the first derivative test to determine the local extrema y = sinx cosx?

Hint: We can solve this question using differentiation. First, we have to differentiate the given equation. After finding the derivative, we have to equate it to the zero value. Then we have to solve the equation to get the required answer. The solution of the equation are the points of local extrema for the given question.

Complete step by step solution:
According to the problem, we are asked to use the first derivative test to determine the local extrema y = sinx cosx.
For this , we first have to differentiate the given equation. After finding the derivative, we have to equate it to the zero value. Then we have to solve the equation to get the required answer. The solution of the equation are the points of local extrema for the given question.
Here, we take the equation given in the question as equation 1.
$ y=\sin x\cos x$ ------ (1)
Then we have to differentiate the equation 1. After differentiating the equation, we get the following:
$ \Rightarrow \dfrac{dy}{dx}=\dfrac{d\sin x\cos x}{dx}$
$ \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{2}\dfrac{d\sin 2x}{dx}$
$ \Rightarrow \dfrac{dy}{dx}=\cos 2x$ ---- (3)
Therefore, now we have to just equate the equation 3 to 0. After equating it, we get the values as:
$ 2x=\dfrac{\pi }{2}+n\pi $
$ \Rightarrow x=\dfrac{\pi }{4}+\dfrac{n\pi }{2}$
Therefore, finally, we get the final answer for the local extrema for the given question as $ x=\dfrac{\pi }{4}+\dfrac{n\pi }{2}$ [Here n is a integer].

Note: You have to know basic differentiation for this. You could also do the differentiation of the given question using the product rule but the simpler way is to use the trigonometric identities. Therefore knowing the trigonometric identities will be helpful.