Question

If $y=\frac{A}{x}+B{x}^2$, then $x^2\left(\frac{d^{2}y}{d{x}^2}\right)$is equal to ?

• A. $2y$
• B. $y^2$
• C. $y^3$
• D. $y^4$

Answer 1

The correct option is A

$2y$

Explanation for the correct option.

Step 1. Find the value of $\frac{dy}{dx}$.

Differentiate the function $y=\frac{A}{x}+B{x}^2$ with respect to $x$.

$\begin{array}{rcl}\frac{dy}{dx}& =& \frac{d}{dx}\left(\frac{A}{x}+B{x}^2\right)\\ & =& -\frac{A}{x^2}+2Bx\end{array}$

Step 2. Find the value of the given expression.

Differentiate the function $\frac{dy}{dx}=-\frac{A}{x^2}+2Bx$ with respect to $x$.

$\begin{array}{rcl}\frac{d^{2}y}{d{x}^2}& =& \frac{d}{dx}\left(-\frac{A}{x^2}+2Bx\right)\\ & =& -\frac{-2A}{x^3}+2B\\ & =& 2\left(\frac{A}{x^3}+B\right)\end{array}$

Now multiply both sides by $x^2$ to get the value of $x^2\left(\frac{d^{2}y}{d{x}^2}\right)$.

$\begin{array}{rcl}{x}^2\left(\frac{d^{2}y}{d{x}^2}\right)& =& x^2\times 2\left(\frac{A}{x^3}+B\right)\\ & =& 2\left(\frac{A}{x}+B{x}^2\right)\\ & =& 2y\left[\because y=\frac{A}{x}+{\mathrm{Bx}}^2\right]\end{array}$

Hence, the correct option is A.