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

$$\frac{dy}{dx}=\frac{1}{y+1}$$ $$\frac{d^2y}{dx^2}=-\frac{1}{(y+1)^3}$$
$$y^2-2x=1-2y$$ 1) Find $dy/dx$: Differentiate both sides of the equation with respect to $x$: $$\frac{d}{dx}(y^2)-\frac{d}{dx}(2x)=\frac{d}{dx}1-\frac{d}{dx}(2y)$$ $$2y\frac{dy}{dx}-2=0-2\frac{dy}{dx}$$ $$y\frac{dy}{dx}-1=-\frac{dy}{dx}$$ Isolate the terms with $dy/dx$ into one side and solve for $dy/dx$: $$y\frac{dy}{dx}+\frac{dy}{dx}=1$$ $$(y+1)\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{y+1}$$ 2) Find $d^2y/dx^2$: Differentiate the derivative $dy/dx$ as normal, using Quotient Rule: $$\frac{d^2y}{dx^2}=\frac{d}{dx}\Big(\frac{1}{y+1}\Big)=\frac{-1\frac{d}{dx}(y+1)}{(y+1)^2}=-\frac{\frac{dy}{dx}}{(y+1)^2}$$ Now we can use the result $dy/dx=1/(y+1)$ from part 1) to do substitution here: $$\frac{d^2y}{dx^2}=-\frac{\frac{1}{y+1}}{(y+1)^2}=-\frac{1}{(y+1)^3}$$