Question

Given \( y = 4e^{-x} - 2e^{-2x} - e^{-3x} \), find \( y'' \).

• A. \( 4e^{-x} - 4e^{-2x} - 3e^{-3x} \)
• B. \( 4e^{-x} - 2e^{-2x} - 6e^{-3x} \)
• C. \( 4e^{-x} - 2e^{-2x} - 3e^{-3x} \)
• D. \( 4e^{-x} - 2e^{-2x} - 5e^{-3x} \)

Hint:

When differentiating exponential functions, remember that the derivative of \( e^{ax} \) is \( ae^{ax} \). This helps in finding higher-order derivatives efficiently.

Answer 1

The Correct Option is C

Step 1: Differentiate the given function.
The given function is: \[ y = 4e^{-x} - 2e^{-2x} - e^{-3x} \] We need to find the second derivative \( y'' \). First, differentiate \( y \) with respect to \( x \) to get \( y' \): \[ y' = \frac{d}{dx} \left( 4e^{-x} \right) - \frac{d}{dx} \left( 2e^{-2x} \right) - \frac{d}{dx} \left( e^{-3x} \right) \] \[ y' = -4e^{-x} + 4e^{-2x} + 3e^{-3x} \]
Step 2: Differentiate again to get \( y'' \).
Now, differentiate \( y' \) to get the second derivative: \[ y'' = \frac{d}{dx} \left( -4e^{-x} \right) + \frac{d}{dx} \left( 4e^{-2x} \right) + \frac{d}{dx} \left( 3e^{-3x} \right) \] \[ y'' = 4e^{-x} - 8e^{-2x} - 9e^{-3x} \]
Final Answer: \( 4e^{-x} - 2e^{-2x} - 3e^{-3x} \).