Question

The second derivative of \( \sin 3x \cos 5x \) is:

• A. \( 2\sin 2x - 16\sin 8x \)
• B. \( 2\sin 2x + 16\sin 8x \)
• C. \( 2\sin 2x - 32\sin 8x \)
• D. \( 2\sin 2x + 32\sin 8x \)

Hint:

Before differentiating products like \( \sin ax \cos bx \), first convert them into sums using product-to-sum identities. This simplifies higher derivatives enormously.

Answer 1

The Correct Option is C

Concept: Whenever a product of trigonometric functions appears in differentiation problems, especially while finding higher order derivatives, it is usually advantageous to simplify the expression first using trigonometric identities. The standard identity used here is: \[ \sin A \cos B = \frac{1}{2} \left[ \sin(A+B)+\sin(A-B) \right] \] This converts the product into a sum, making differentiation much easier and more systematic.

Step 1: Convert the product into a sum using trigonometric identities. We are given: \[ y=\sin 3x \cos 5x \] Using the identity: \[ \sin A \cos B = \frac12 \left[ \sin(A+B)+\sin(A-B) \right] \] Taking \[ A=3x, \qquad B=5x \] we obtain: \[ y = \frac12 \left[ \sin(3x+5x)+\sin(3x-5x) \right] \] \[ y = \frac12 \left[ \sin 8x+\sin(-2x) \right] \] Now using the property: \[ \sin(-\theta)=-\sin\theta \] we get: \[ y = \frac12 \left[ \sin 8x-\sin 2x \right] \] Hence, \[ y = \frac12\sin 8x - \frac12\sin 2x \]

Step 2: Find the first derivative. Differentiate term-by-term. Using: \[ \frac{d}{dx}(\sin ax)=a\cos ax \] we get: \[ \frac{dy}{dx} = \frac12(8\cos 8x) - \frac12(2\cos 2x) \] \[ \frac{dy}{dx} = 4\cos 8x-\cos 2x \] Thus, the first derivative is: \[ \boxed{ \frac{dy}{dx}=4\cos 8x-\cos 2x } \]

Step 3: Differentiate again to obtain the second derivative. Now differentiate: \[ \frac{dy}{dx}=4\cos 8x-\cos 2x \] Using: \[ \frac{d}{dx}(\cos ax)=-a\sin ax \] we get: \[ \frac{d^2y}{dx^2} = 4(-8\sin 8x)-(-2\sin 2x) \] \[ \frac{d^2y}{dx^2} = -32\sin 8x+2\sin 2x \] Rearranging the terms, \[ \boxed{ \frac{d^2y}{dx^2} = 2\sin 2x-32\sin 8x } \] Hence, the correct answer is: \[ \boxed{(C)\ 2\sin 2x-32\sin 8x} \]