## Calculus: Early Transcendentals (2nd Edition)

$y''=-4\sin\theta\cos\theta$
$y=\cos\theta\sin\theta$ 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.