Question

Find (d^2 y) / (d x^2) if x^2 y^2 = 1.

Answer 1

Answer:  $\displaystyle \frac{2y}{x^2}$ which is choice A

========================================================

Work Shown

Apply the derivative to both sides with respect to x.

$x^2y^2 = 1\\\\\frac{d}{dx}(x^2y^2) = \frac{d}{dx}(1)\\\\\frac{d}{dx}(x^2)y^2+x^2\frac{d}{dx}(y^2) = 0 \ \text{ ... product rule}\\\\2xy^2+x^2 2y \frac{dy}{dx} = 0 \ \text{ ... chain rule}$

Let's isolate dy/dx

$2xy^2+x^2 2y \frac{dy}{dx} = 0\\\\2x^2y \frac{dy}{dx} = -2xy^2\\\\\frac{dy}{dx} = \frac{-2xy^2}{2x^2y}\\\\\frac{dy}{dx} = -\frac{y}{x}$

Apply another derivative.

$\frac{dy}{dx} = -\frac{y}{x}\\\\\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(-\frac{y}{x}\right)\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-y x^{-1}\right)\\\\\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-y\right) x^{-1}-y\frac{d}{dx}\left(x^{-1}\right)\\\\\frac{d^2y}{dx^2} = -\frac{dy}{dx} x^{-1}+yx^{-2}$

Then substitute in what we found for the 1st derivative and simplify.

$\frac{d^2y}{dx^2} = -\frac{dy}{dx} x^{-1}+yx^{-2}\\\\\frac{d^2y}{dx^2} = -\left(-\frac{y}{x}\right) *\frac{1}{x}+y*\frac{1}{x^2}\\\\\frac{d^2y}{dx^2} = \frac{y}{x^2}+\frac{y}{x^2}\\\\\frac{d^2y}{dx^2} = \frac{y+y}{x^2}\\\\\frac{d^2y}{dx^2} = \frac{2y}{x^2}$