## Calculus 8th Edition

a)$y'=\frac{2y^2}{x^2}$ b) $y'=\frac{1}{2(1-2x)^2}$ c) See below
a) $\frac{d}{dx}(2x^{-1}-y^{-1}=4)\\ -2x^{-2}+y^{-2}y'=0\\ -\frac{2}{x^2}+\frac{y'}{y^2}=0\\ \frac{y'}{y^2}=\frac{2}{x^2}\\ y'=\frac{2y^2}{x^2}$ b) First we have to solve for $y$. $\frac{2}{x}-\frac{1}{y}=4\\ \frac{2}{x}-4=\frac{1}{y}\\ y=\frac{1}{\frac{2}{x}-4}\\ y=\frac{1}{\frac{2-4x}{x}}\\ y=\frac{x}{2-4x}\\$ Then we take the derivative of $y$ by using the Quotient Rule. $y'=\frac{(2-4x)(1)-(x)(-4)}{(2-4x)^2}\\ y'=\frac{2}{(2-4x)^2}\\ y'=\frac{2}{(2(1-2x))^2}\\ y'=\frac{2}{4(1-2x)^2}\\ y'=\frac{1}{2(1-2x)^2}\\$ c) We plug in $y=\frac{x}{2-4x}$ into a). $y'=\frac{{2(\frac{x}{2-4x})}^2}{x^2}\\ y'=\frac{\frac{2{x^2}}{{(2-4x)^2}}}{x^2}\\ y'=\frac{2x^2}{x^2(2-4x)^2}\\ y'=\frac{2}{(2-4x)^2}\\ y'=\frac{2}{(2(1-2x))^2}\\ y'=\frac{2}{4(1-2x)^2}\\ y'=\frac{1}{2(1-2x)^2}\\$ Therefore, our solutions to parts a) and b) are consistent.