Let \(y = xe^{x}\). Consider the following statements:
(A) \(\displaystyle\frac{dy}{dx} = (x + 1)e^{x}\)
(B) \(\displaystyle\frac{d^{2}y}{dx^{2}} = (x + 2)e^{x}\)
(C) \(\displaystyle\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + y = 0\)
(D) The value of \(\displaystyle\frac{dy}{dx}\) at \(x = 0\) is 0.
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → (A), (B) and (C) only
By the product rule:
\( \displaystyle \frac{dy}{dx} = e^{x} + xe^{x} = (x + 1)e^{x} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = e^{x} + (x + 1)e^{x} = (x + 2)e^{x} \)
So \(y' = (x + 1)e^{x}\), \(y'' = (x + 2)e^{x}\) and \(y'(0) = 1\).
Checking the statements
(A) \(\frac{dy}{dx} = (x + 1)e^{x}\) — Correct. Product rule: \(e^{x} + xe^{x}\).
(B) \(\frac{d^{2}y}{dx^{2}} = (x + 2)e^{x}\) — Correct. Derivative of \((x + 1)e^{x}\) is \(e^{x} + (x + 1)e^{x}\).
(C) \(\frac{d^{2}y}{dx^{2}} - 2\frac{dy}{dx} + y = 0\) — Correct. \(e^{x}[(x + 2) - 2(x + 1) + x] = 0\).
(D) The value of \(\frac{dy}{dx}\) at \(x = 0\) is 0. — Incorrect. \(y'(0) = (0 + 1)e^{0} = 1\).