Calculus: Early Transcendentals 8th Edition

(a) First, carry out the derivative process using the Product Rule as usual. Then, use the Chain Rule for any derivatives that cannot be dealt with in the normal way. Finally, do the math, simplify using the formula $\cos(a+b)=\cos a\cos b-\sin a\sin b$ (b) $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos^{n-1}x\sin(n+1)x$$
(a) $$\frac{d}{dx}(\sin^n x\cos nx)$$ $$=\cos nx\frac{d}{dx}(\sin^n x)+\sin^n x\frac{d}{dx}(\cos nx)$$ *First, consider $\frac{d}{dx}(\sin^n x)$ $$\frac{d}{dx}(\sin^n x)=\frac{d(\sin^n x)}{d(\sin x)}\frac{d(\sin x)}{dx}$$ $$\frac{d}{dx}(\sin^n x)=n\sin^{n-1}x\cos x$$ *Second, consider $\frac{d}{dx}(\cos nx)$ $$\frac{d}{dx}(\cos nx)=\frac{d(\cos nx)}{d(nx)}\frac{ndx}{dx}$$ $$\frac{d}{dx}(\cos nx)=-n\sin nx$$ Therefore, $$\frac{d}{dx}(\sin^n x\cos nx)=\cos nx(n\sin^{n-1}x\cos x)+\sin^n x(-n\sin nx)$$ $$\frac{d}{dx}(\sin^n x\cos nx)=n\cos nx\sin^{n-1}x\cos x-n\sin^n x\sin nx$$ $$\frac{d}{dx}(\sin^n x\cos nx)=n\sin^{n-1}x(\cos x\cos nx-\sin x\sin nx)$$ We know that $\cos(a+b)=\cos a\cos b-\sin a\sin b$ Therefore, $\cos x\cos nx-\sin x\sin nx=\cos(x+nx)=\cos (n+1)x$ So, $$\frac{d}{dx}(\sin^n x\cos nx)=n\sin^{n-1}x\cos(n+1)x$$ The formula is proven. (b) $$\frac{d}{dx}(\cos^n x\cos nx)$$ $$=\cos nx\frac{d}{dx}(\cos^n x)+\cos^n x\frac{d}{dx}(\cos nx)$$ *First, consider $\frac{d}{dx}(\cos^n x)$ $$\frac{d}{dx}(\cos^n x)=\frac{d(\cos^n x)}{d(\cos x)}\frac{d(\cos x)}{dx}$$ $$\frac{d}{dx}(\cos^n x)=-n\cos^{n-1}x\sin x$$ *Second, consider $\frac{d}{dx}(\cos nx)$ Yet, we already know from part a) that $\frac{d}{dx}(\cos nx)=-n\sin nx$ Therefore, $$\frac{d}{dx}(\cos^n x\cos nx)=\cos nx(-n\cos^{n-1}x\sin x)+\cos^n x(-n\sin nx)$$ $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos nx\cos^{n-1}x\sin x-n\cos^n x\sin nx$$ $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos^{n-1}x(\sin x\cos nx+\cos x\sin nx)$$ We know that $\sin(a+b)=\sin a\cos b+\cos a\sin b$ Therefore, $\sin x\cos nx+\cos x\sin nx=\sin(x+nx)=\sin (n+1)x$ So, $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos^{n-1}x\sin(n+1)x$$