Question

For $\text{n} \in \mathbb{N}$ if $y = \text{a}x^{\text{n}+1} + \text{b}x^{-\text{n}}$, then $x^2 \frac{\text{d}^2 y}{\text{d}x^2} =$

• A. $\text{n}(\text{n} - 1)y$
• B. $(\text{n} - 1)y$
• C. $\text{n}(\text{n} + 1)y$
• D. $(\text{n} + 1)y$

Hint:

Look for factorization after differentiation.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
We need to find the second derivative of a given polynomial-like function with positive and negative exponents, multiply it by $x^2$, and then express the final result back in terms of the original function $y$. Step 2: Key Formula or Approach:
Use the standard power rule for differentiation: $\frac{d}{dx} (x^k) = k x^{k-1}$. Apply this twice to find $\frac{d^2y}{dx^2}$, multiply by $x^2$, and factor out constants to reconstruct the expression for $y$. Step 3: Detailed Explanation:
Given the function: \[ y = a x^{n+1} + b x^{-n} \] Find the first derivative $\frac{dy}{dx}$ using the power rule: \[ \frac{dy}{dx} = a(n+1) x^{(n+1)-1} + b(-n) x^{-n-1} \] \[ \frac{dy}{dx} = a(n+1) x^n - bn x^{-n-1} \] Now, find the second derivative $\frac{d^2y}{dx^2}$ by differentiating again: \[ \frac{d^2y}{dx^2} = a(n+1)(n) x^{n-1} - bn(-n-1) x^{-n-1-1} \] \[ \frac{d^2y}{dx^2} = an(n+1) x^{n-1} + bn(n+1) x^{-n-2} \] The question asks for $x^2 \frac{d^2y}{dx^2}$. Multiply the entire expression by $x^2$: \[ x^2 \frac{d^2y}{dx^2} = x^2 \left( an(n+1) x^{n-1} + bn(n+1) x^{-n-2} \right) \] Distribute $x^2$ inside the parentheses. Remember to add exponents: $x^2 \cdot x^{n-1} = x^{n+1}$ and $x^2 \cdot x^{-n-2} = x^{-n}$. \[ x^2 \frac{d^2y}{dx^2} = an(n+1) x^{n+1} + bn(n+1) x^{-n} \] Notice that both terms contain the common scalar factor $n(n+1)$. Factor it out: \[ x^2 \frac{d^2y}{dx^2} = n(n+1) \left( a x^{n+1} + b x^{-n} \right) \] Observe that the expression inside the parentheses is exactly our original function $y$: \[ x^2 \frac{d^2y}{dx^2} = n(n+1) y \] Step 4: Final Answer:
The expression equals $\text{n}(\text{n} + 1)y$.