If \( y=ax^{n+1}+bx^{-n} \), then \( x^{2}\displaystyle\frac{d^{2}y}{dx^{2}} \) is equal to
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( n(n+1)y \)
By the power rule,
\( \displaystyle \frac{dy}{dx}=a(n+1)x^{n}-nbx^{-n-1} \)
\( \displaystyle \frac{d^{2}y}{dx^{2}}=a(n+1)n\,x^{n-1}+n(n+1)b\,x^{-n-2} \)
\( \displaystyle =n(n+1)\left[ax^{n-1}+bx^{-n-2}\right] \)
Multiplying by \( x^{2} \):
\( \displaystyle x^{2}\frac{d^{2}y}{dx^{2}}=n(n+1)\left[ax^{n+1}+bx^{-n}\right] \)
\( \displaystyle =n(n+1)y \)
Shortcut: Each term \( x^{m} \) gives
\( \displaystyle x^{2}\frac{d^{2}}{dx^{2}}\left(x^{m}\right)=m(m-1)x^{m} \)
For m = n + 1 this factor is (n + 1)n, and for m = −n it is (−n)(−n − 1) = n(n + 1). Both terms get the same factor, so the answer is n(n + 1)y.