## Calculus: Early Transcendentals 8th Edition

$g''(\pi/6)=\sqrt{3}-\dfrac{\pi}{12}\approx1.470$
If $g(\theta)=\theta\sin\theta$, find $g''(\pi/6)$ Find the first derivative by using the product rule: $g'(\theta)=\theta(\sin\theta)'+(\theta)'\sin\theta=...$ $...=\theta\cos\theta+\sin\theta$ Use the product rule one more time to evaluate the second derivative: $g''(\theta)=\theta(\cos\theta)'+(\theta)'\cos\theta+\cos\theta=...$ $...=-\theta\sin\theta+\cos\theta+\cos\theta=2\cos\theta-\theta\sin\theta$ Substitute $\theta$ by $\pi/6$ in $g''(\theta)$ and simplify to find $g''(\pi/6)$: $g''(\pi/6)=2\cos\dfrac{\pi}{6}-\dfrac{\pi}{6}\sin\dfrac{\pi}{6}=2\Big(\dfrac{\sqrt{3}}{2}\Big)-\Big(\dfrac{\pi}{6}\Big)\Big(\dfrac{1}{2}\Big)=...$ $...=\sqrt{3}-\dfrac{\pi}{12}\approx1.470$