Answer
$y''=-4\sin\theta\cos\theta$
Work Step by Step
$y=\cos\theta\sin\theta$
Apply the product rule to find the first derivative:
$y'=\cos\theta(\sin\theta)'+\sin\theta(\cos\theta)'=...$
Evaluate the derivatives indicated and simplify:
$...=(\cos\theta)(\cos\theta)+(\sin\theta)(-\sin\theta)=\cos^{2}\theta-\sin^{2}\theta$
The expression above is the first derivative.
Evaluate the second derivative by differentiating the first derivative found:
$y''=2\cos\theta(\cos\theta)'-2\sin\theta(\sin\theta)'=...$
Evaluate the derivatives indicated and simplify:
$...=2(\cos\theta)(-\sin\theta)-2(\sin\theta)(\cos\theta)=...$
$...=-2\sin\theta\cos\theta-2\sin\theta\cos\theta=-4\sin\theta\cos\theta$
The expression above is the second derivative.
