Answer is: option3

Solution:

Using the product rule: \[ \frac{d}{dx} (xe^x) = x \frac{d}{dx} (e^x) + e^x \frac{d}{dx} (x) \] \[ = x e^x + e^x \] \[ = (x + 1) e^x \] Again using the product rule: \[ \frac{d^2 y}{dx^2} = \frac{d}{dx} [(x + 1) e^x] \] \[ = (x + 1) \frac{d}{dx} (e^x) + e^x \frac{d}{dx} (x + 1) \] \[ = (x + 1) e^x + e^x \] \[ = (x + 2) e^x \] We observe the pattern: \[ \frac{d^1 y}{dx} = (x + 1) e^x \] \[ \frac{d^2 y}{dx^2} = (x + 2) e^x \] From this, we generalize: \[ \frac{d^n y}{dx^n} = (x + n) e^x \]