Answer
a) $y'=\frac{2y^2}{x^2}$
b) $y'=\frac{1}{2(1-2x)^2}$
c) See below
Work Step by Step
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.
