Find the derivative of $sinx$ $cosx$

The correct answer is $cos2x$

Let f(x) = $sinx$ $cosx$

Using the Product rule,

$f'(x)$ = $sinx \dfrac{d}{dx}cosx$ + $cosx \dfrac{d}{dx}sinx$

= $sinx (-sinx)$ + $cosx (cosx)$

= $-sin^2x$ + $cos^2x$

= $cos^2x - sin^2x$

= $cos2x$