If \( y=ax^{n+2}+\displaystyle\frac{b}{x^{n+1}} \), then \( x^2\displaystyle\frac{d^2y}{dx^2} \) is equal to
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (n+1)(n+2)y \)
Writing \( y=ax^{n+2}+bx^{-(n+1)} \), the power rule gives
\( \displaystyle \frac{dy}{dx}=a(n+2)x^{n+1}-b(n+1)x^{-(n+2)} \)
\( \displaystyle \frac{d^2y}{dx^2}=a(n+2)(n+1)x^{n}+b(n+1)(n+2)x^{-(n+3)} \)
Multiplying by \( x^2 \):
\( \displaystyle x^2\frac{d^2y}{dx^2}=(n+1)(n+2)\left[ax^{n+2}+bx^{-(n+1)}\right] \)
\( \displaystyle =(n+1)(n+2)y \)