## Calculus: Early Transcendentals 8th Edition

$$y'=-3\sin (\sin3\theta)\cos3\theta$$ $$y''=9\sin(\sin3\theta)\sin3\theta-9\cos(\sin3\theta)\cos^23\theta$$
$$y=\cos(\sin3\theta)$$ 1) Find $y'$ According to Chain Rule, $$y'=\frac{dy}{d\theta}=\frac{d(\cos(\sin3\theta))}{d(\sin3\theta)}\frac{d(\sin(3\theta))}{d\theta}$$ $$y'=-\sin (\sin3\theta)\frac{d(\sin(3\theta))}{d\theta}$$ According to Chain Rule again, $$y'=-\sin (\sin3\theta)\frac{d(\sin(3\theta))}{d(3\theta)}\frac{d(3\theta)}{d\theta}$$ $$y'=-\sin (\sin3\theta)\cos3\theta\times(3\times1)$$ $$y'=-3\sin (\sin3\theta)\cos3\theta$$ 2) Find $y''$ $$y''=-3[(\sin(\sin3\theta))'\cos3\theta+\sin(\sin3\theta)(\cos3\theta)']$$ $$y''=-3[A\cos3\theta+\sin(\sin3\theta)B]$$ *Consider A: $$A=(\sin(\sin3\theta))'$$ $$A=\frac{d(\sin(\sin3\theta))}{d(\sin3\theta)}\frac{d(\sin3\theta)}{d(3\theta)}\frac{d(3\theta)}{d(\theta)}$$ $$A=\cos(\sin(3\theta))\cos3\theta\times3$$ $$A=3\cos(\sin(3\theta))\cos3\theta$$ *Consider B: $$B=(\cos3\theta)'$$ $$B=\frac{d(\cos3\theta)}{d(3\theta)}\frac{d(3\theta)}{d(\theta)}$$ $$B=-\sin3\theta\times3$$ $$B=-3\sin3\theta$$ Therefore, $$y''=-3[(3\cos(\sin(3\theta))\cos3\theta)\cos3\theta+\sin(\sin3\theta)(-3\sin3\theta)]$$ $$y''=-3[3\cos(\sin(3\theta))\cos^23\theta-3\sin(\sin3\theta)\sin3\theta]$$ $$y''=9\sin(\sin3\theta)\sin3\theta-9\cos(\sin3\theta)\cos^23\theta$$