If \(y = e^{3x} + e^{-3x}\), then \(\displaystyle\frac{d^{2}y}{dx^{2}}\) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(9y\)
Since \(\frac{d}{dx}e^{kx} = k\,e^{kx}\), each differentiation brings down the factor k:
\( \displaystyle \frac{dy}{dx} = 3e^{3x} - 3e^{-3x} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}} = 9e^{3x} + 9e^{-3x} \)
\( = 9\left(e^{3x} + e^{-3x}\right) \)
\( = 9y \)
The sign comes back positive in the second derivative because \((-3)^{2} = 9\).