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


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

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.