## Thomas' Calculus 13th Edition

a. Given $x^2-y^2=1$, we have $2xdx-2ydy=0$, which gives $\frac{dy}{dx}=\frac{x}{y}=xy^{-1}$ b. Differentiating against $x$ again, we have $\frac{d^2y}{dx^2}=y^{-1}-xy^{-2}\frac{dy}{dx}=y^{-1}-xy^{-2}(xy^{-1})=y^{-1}-x^2y^{-3}=\frac{y^2-x^2}{y^3}=-\frac{1}{y^3}$