Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 95

Answer

(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) $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos^{n-1}x\sin(n+1)x$$

Work Step by Step

(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, $$\frac{d}{dx}(\cos^n x\cos nx)=-n\cos^{n-1}x\sin(n+1)x$$
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