**Calculus Problem: Derivative of Trigonometric Function****Problem Statement:**Find \(\frac{dy}{dx}\) given \( \sin(3x)\cos(x) \).**Solution Steps:**To find \(\frac{dy}{dx}\) for the given function \(y = \sin(3x)\cos(x)\), we can utilize the product rule for differentiation. The product rule states that if \(u(x)\) and \(v(x)\) are functions of \(x\), then the derivative of their product is given by:\[\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)\]Let \(u(x) = \sin(3x)\) and \(v(x) = \cos(x)\).First, compute the derivatives of each function:1. \(u(x) = \sin(3x)\) \(u'(x) = \frac{d}{dx}[\sin(3x)] = 3\cos(3x)\) (using the chain rule)2. \(v(x) = \cos(x)\) \(v'(x) = \frac{d}{dx}[\cos(x)] = -\sin(x)\)Now apply the product rule:\[\frac{dy}{dx} = u'(x)v(x) + u(x)v'(x)\]Substitute \(u'(x)\), \(u(x)\), \(v(x)\), and \(v'(x)\):\[\frac{dy}{dx} = [3\cos(3x)] \cos(x) + [\sin(3x)] [-\sin(x)] \]Simplify the expression:\[\frac{dy}{dx} = 3\cos(3x)\cos(x) - \sin(3x)\sin(x)\]Therefore, the derivative of \(y = \sin(3x)\cos(x)\) is:\[\frac{dy}{dx} = 3\cos(3x)\cos(x) - \sin(3x)\sin(x)\]This is the required derivative.

Name: **Calculus Problem: Derivative of Trigonometric Function****Problem S
Uploaded: 2024-03-20T06:46:42.000Z
Channel: Bartleby
Description: **Calculus Problem: Derivative of Trigonometric Function****Problem Statement:**Find \(\frac{dy}{dx}\) given \( \sin(3x)\cos(x) \).**Solution Steps:**To find \(\frac{dy}{dx}\) for the given function \(y = \sin(3x)\cos(x)\), we can utilize the produc