## Calculus 10th Edition

$\dfrac{d^2y}{dx^2}=\dfrac{6xy-16}{x^3}$.
First Derivative: $\dfrac{d}{dx}(x^2y)-\dfrac{d}{dx}(4x)=\dfrac{d}{dx}(5)\rightarrow$ $2xy+\dfrac{dy}{dx}(x^2)-4=0\rightarrow$ $\dfrac{dy}{dx}=\dfrac{4-2xy}{x^2}$ Second Derivative: $\dfrac{d}{dx}(\dfrac{dy}{dx})=\dfrac{d}{dx}(\dfrac{4-2xy}{x^2})\rightarrow$ Using the Quotient Rule: $\dfrac{d^2y}{dx^2}=\dfrac{(-2y+\dfrac{dy}{dx}(-2x))(x^2)-(2x)(4-2xy)}{x^4}\rightarrow$ $\dfrac{2x^2y-2x^3(\dfrac{4-2xy}{x^2})-8x}{x^4}\rightarrow$ $\dfrac{6xy-16}{x^3}.$