## Calculus: Early Transcendentals (2nd Edition)

$\frac{d^2y}{dx^2} = \frac{x\sin(x)+4\cos(x)+6xy}{x^3}$
$sin(x)+x^2y = 10$ First Derivative $cos(x)+2xy+x^2(\frac{dy}{dx}) = 0$ $\frac{dy}{dx} = \frac{-cos(x)-2xy}{x^2} = \frac{-cos(x)}{x^2} - \frac{2xy}{x^2} = \frac{-cos(x)}{x^2}-\frac{2y}{x}$ Second Derivative: $\frac{d^2y}{dx^2} = \frac{(sin(x))(x^2)-(2x)(-cos(x))}{x^4} - \frac{(2\frac{dy}{dx})(x)-(1)(2y)}{x^2} = \frac{xsin(x)+2cos(x)}{x^3} - \frac{\frac{-2cos(x)}{x}-2y-4y}{x^2} = \frac{x\sin(x)+4\cos(x)+6xy}{x^3}$