## University Calculus: Early Transcendentals (3rd Edition)

a) $$\frac{d^2y}{dx^2}=\frac{-2x^4-2xy^3}{y^5}$$ b) $$\frac{d^2y}{dx^2}=\frac{-2xy^2-1}{x^4y^3}$$
a) $$x^3+y^3=1$$ 1) First, we would find $dy/dx$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$3x^2+3y^2\frac{dy}{dx}=0$$ $$3y^2\frac{dy}{dx}=-3x^2$$ $$\frac{dy}{dx}=-\frac{3x^2}{3y^2}=-\frac{x^2}{y^2}$$ 2) Next, we use implicit differentiation again on $dy/dx$ to find $d^2y/dx^2$: $$\frac{d^2y}{dx^2}=-\Big(\frac{(x^2)'y^2-x^2(y^2)'}{y^4}\Big)=-\Big(\frac{2xy^2-x^2\times2y\times\frac{dy}{dx}}{y^4}\Big)$$ $$\frac{d^2y}{dx^2}=-\Big(\frac{2xy^2-2x^2y\frac{dy}{dx}}{y^4}\Big)=\frac{2x^2y\frac{dy}{dx}-2xy^2}{y^4}$$ Here, we replace $dy/dx=-(x^2/y^2)$: $$\frac{d^2y}{dx^2}=\frac{2x^2y(-\frac{x^2}{y^2})-2xy^2}{y^4}=\frac{-\frac{2x^4}{y}-2xy^2}{y^4}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{-2x^4-2xy^3}{y}}{y^4}=\frac{-2x^4-2xy^3}{y^5}$$ b) $$y^2=1-\frac{2}{x}$$ 1) First, we would find $dy/dx$ Using implicit differentiation, we differentiate both sides of the equation with respect to $x$: $$2y\frac{dy}{dx}=0-\Big(\frac{(2)'x-2x'}{x^2}\Big)=-\Big(\frac{0\times x-2}{x^2}\Big)$$ $$2y\frac{dy}{dx}=-\Big(\frac{-2}{x^2}\Big)=\frac{2}{x^2}$$ $$\frac{dy}{dx}=\frac{2}{x^2\times2y}=\frac{1}{x^2y}$$ 2) Next, we use implicit differentiation again on $dy/dx$ to find $d^2y/dx^2$: $$\frac{d^2y}{dx^2}=\frac{\frac{d(1)}{dx}x^2y-1\times\frac{d(x^2y)}{dx}}{(x^2y)^2}=\frac{0\times x^2y-(2xy+x^2\frac{dy}{dx})}{x^4y^2}$$ $$\frac{d^2y}{dx^2}=\frac{-2xy-x^2\frac{dy}{dx}}{x^4y^2}$$ Here, we replace $dy/dx=1/(x^2y)$: $$\frac{d^2y}{dx^2}=\frac{-2xy-\frac{x^2}{x^2y}}{x^4y^2}=\frac{-2xy-\frac{1}{y}}{x^4y^2}$$ $$\frac{d^2y}{dx^2}=\frac{\frac{-2xy^2-1}{y}}{x^4y^2}=\frac{-2xy^2-1}{x^4y^3}$$