Question

If $y=a{x}^{n+1}+b{x}^{-n}$, then $x^2\frac{d^{2}y}{d{x}^2}=$

• A. $n\left(n-1\right)y$
• B. $n\left(n+1\right)y$
• C. $ny$
• D. $n^{2}y$

Answer 1

The correct option is B

$n\left(n+1\right)y$

Explanation for the correct option.

Step 1. Differentiate the equation with respect to $x$.

Differentiate the equation $y=a{x}^{n+1}+b{x}^{-n}$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(a{x}^{n+1}+b{x}^{-n}\right)\\ & =& a\left(n+1\right)x^{n+1-1}+b\left(-n\right)x^{-n-1}\\ & =& a\left(n+1\right)x^n-bn{x}^{-\left(n+1\right)}\end{array}$

Step 2. Find the value of $x^2\frac{d^{2}y}{d{x}^2}$.

Differentiate $\frac{dy}{dx}=a\left(n+1\right)x^n-bn{x}^{-\left(n+1\right)}$ with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left(\frac{dy}{dx}\right)=\frac{d}{dx}\left(a\left(n+1\right)x^n-bn{x}^{-\left(n+1\right)}\right) \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=a\left(n+1\right)n{x}^{n-1}-bn\left[-\left(n+1\right)\right]x^{-n-1-1} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=an(n+1)x^{n-1}+bn(n+1)x^{-n-2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}=n\left(n+1\right)\left[a{x}^{n-1}+b{x}^{-n-2}\right] \end{gathered}$$

Now, multiply both sides by $x^2$.

$\begin{array}{rcl}{x}^2\frac{d^{2}y}{d{x}^2}& =& x^{2}n\left(n+1\right)\left[a{x}^{n-1}+b{x}^{-n-2}\right]\\ & =& n(n+1)\left[a{x}^2{x}^{n-1}+b{x}^2{x}^{-n-2}\right]\\ & =& n\left(n+1\right)\left[a{x}^{2+n-1}+b{x}^{2-n-2}\right]\left[a^m{a}^n=a^{m+n}\right]\\ & =& n(n+1)\left[a{x}^{n+1}+b{x}^{-n}\right]\\ & =& n(n+1)y\left[y=a{x}^{n+1}+b{x}^{-n}\right]\end{array}$

Thus the value of $x^2\frac{d^{2}y}{d{x}^2}$ is $n\left(n+1\right)y$.

Hence, the correct option is B.