If \(y = 500e^{7x} + 600e^{-7x}\), then \(\displaystyle\frac{d^2y}{dx^2}\) is equal to:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 49y
Since \(\frac{d}{dx}e^{kx} = ke^{kx}\), each differentiation brings down a factor 7 from the first term and −7 from the second.
\(\displaystyle\frac{dy}{dx} = 500 \times 7e^{7x} + 600 \times (-7)e^{-7x}\)
\(\displaystyle = 3500e^{7x} - 4200e^{-7x}\)
Differentiating again:
\(\displaystyle\frac{d^2y}{dx^2} = 3500 \times 7e^{7x} - 4200 \times (-7)e^{-7x}\)
\(= 24500e^{7x} + 29400e^{-7x}\)
\(\displaystyle = 49\left(500e^{7x} + 600e^{-7x}\right)\)
\(= 49y\)
NCERT (2026-27 edition):
Second Order Derivative
