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sin x 8x y = 8x sin x dy 8. CoS X 8. %3D dx cos X dy dx x cosx + sin x 8 sin x + 8x cosx 8x2 sin2x dv sin x-x cS X 8x cos x - 8 sin x dx 64x2 sin2x dy x cos - sin x 8 sin x-8x cosx dx 8x2 sinx

sin x 8x y = 8x sin x dy 8. CoS X 8. %3D dx cos X dy dx x cosx + sin x 8 sin x + 8x cosx 8x2 sin2x dv sin x-x cS X 8x cos x - 8 sin x dx 64x2 sin2x dy x cos - sin x 8 sin x-8x cosx dx 8x2 sinx

Calculus: Early Transcendentals
Calculus: Early Transcendentals
8th Edition
ISBN:9781285741550
Author:James Stewart
Publisher:James Stewart
Chapter1: Functions And Models
Section: Chapter Questions
Problem 1RCC: (a) What is a function? What are its domain and range? (b) What is the graph of a function? (c) How...
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**Find the derivative.** Given: \[y = \frac{\sin x}{8x} + \frac{8x}{\sin x}\] **Options:** 1. \(\frac{dv}{dx} = \frac{\cos x}{8} + \frac{8}{\cos x}\) 2. \(\frac{dv}{dx} = \frac{x \cos x + \sin x}{8x^2} + \frac{8 \sin x + 8x \cos x}{\sin^2 x}\) 3. \(\frac{dv}{dx} = \frac{\sin x - x \cos x}{64x^2} + \frac{8x \cos x - 8 \sin x}{\sin^2 x}\) 4. \(\frac{dv}{dx} = \frac{x \cos x - \sin x}{8x^2} + \frac{8 \sin x - 8x \cos x}{\sin^2 x}\)
Transcribed Image Text:**Find the derivative.** Given: \[y = \frac{\sin x}{8x} + \frac{8x}{\sin x}\] **Options:** 1. \(\frac{dv}{dx} = \frac{\cos x}{8} + \frac{8}{\cos x}\) 2. \(\frac{dv}{dx} = \frac{x \cos x + \sin x}{8x^2} + \frac{8 \sin x + 8x \cos x}{\sin^2 x}\) 3. \(\frac{dv}{dx} = \frac{\sin x - x \cos x}{64x^2} + \frac{8x \cos x - 8 \sin x}{\sin^2 x}\) 4. \(\frac{dv}{dx} = \frac{x \cos x - \sin x}{8x^2} + \frac{8 \sin x - 8x \cos x}{\sin^2 x}\)
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