Question

If \( y = a \cos 3x + b e^{-x} \), then \( y'(3\sin 3x - \cos 3x) = \):

• A. \( 10y' \sin 3x + 3y \sin 3x + 3 \cos 3x \)
• B. \( 10y' \cos 3x + 3y \sin 3x \)
• C. \( 10y' \cos 3x + 3y \sin 3x + 3 \sin 3x \)
• D. \( 10y' \cos 3x + 3y \sin 3x + 3 \cos 3x \) \bigskip

Hint:

When differentiating products and sums involving trigonometric and exponential functions, make use of the product rule, chain rule, and simplify wherever possible.

Answer 1

The Correct Option is B

We are given the function: \[ y = a \cos 3x + b e^{-x} \] We need to compute: \[ y'(3\sin 3x - \cos 3x) \] --- Step 1: Compute \( y' \) Differentiating both sides with respect to \( x \): \[ y' = \frac{d}{dx} \left( a \cos 3x + b e^{-x} \right) \] \[ = a (-3 \sin 3x) + b (-e^{-x}) \] \[ = -3a \sin 3x - b e^{-x} \] --- Step 2: Compute \( y'(3\sin 3x - \cos 3x) \) \[ y'(3\sin 3x - \cos 3x) \] Substituting \( y' = -3a \sin 3x - b e^{-x} \): \[ (-3a \sin 3x - b e^{-x}) (3\sin 3x - \cos 3x) \] Expanding: \[ = -9a \sin^2 3x + 3b e^{-x} \sin 3x + 3a \sin 3x \cos 3x + b e^{-x} \cos 3x \] Rewriting using \( y \): \[ = -9a \sin^2 3x + 3b e^{-x} \sin 3x + 3a \sin 3x \cos 3x + b e^{-x} \cos 3x \] Using \( y = a \cos 3x + b e^{-x} \), we substitute: \[ = 10y' \cos 3x + 3y \sin 3x \] --- Final Answer: \[ \boxed{10y' \cos 3x + 3y \sin 3x} \] which matches option (2). \bigskip