Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 3

a.) Determine $y'$ by Implicit Differentiation.
b.) Find the equation explicitly for $y$ and differentiate to get $y'$ in terms of $x$.
c.) Check that your solutions to part (a) and (b) are consistent by substituting the expression for $y$ into your solution for part (a).

a.) Given: $\displaystyle \frac{1}{x} + \frac{1}{y} = 1$
$\displaystyle \frac{d}{dx} \left( \frac{1}{x}\right) + \frac{d}{dx} \left( \frac{1}{y}\right) = \frac{d}{dx} (1)$
$\displaystyle \frac{\left[ (x) \frac{d}{dx} (1) - (1) \frac{d}{dx}(x) \right]}{x^2} + \frac{\left[ (y) \frac{d}{dx} (1) - (1) \frac{d}{dx}(y) \right]}{y^2} =0 $
$\displaystyle \frac{(x)(0)-(1)(1)}{x^2}+\frac{(y)(0)-(1)\left(\frac{dy}{dx}\right)}{y^2} = 0$


$
\begin{equation}
\begin{aligned}
\frac{-1}{x^2} - \frac{\frac{dy}{dx}}{y^2} &= 0 \\
\\
\frac{\frac{dy}{dx}}{y^2} &= \frac{-1}{x^2}\\
\\
\frac{dy}{dx} &= \frac{-y^2}{x^2} \qquad \text{ or } \qquad y' = \frac{-y^2}{x^2} && \text{(Equation 1)}
\end{aligned}
\end{equation}
$


b.) Solving for $y$


$
\begin{equation}
\begin{aligned}
\frac{1}{y} &= 1 - \frac{1}{x}\\
\\
\frac{1}{y} &= \frac{x-1}{x}\\
\\
\frac{y(\cancel{x-1})}{(\cancel{x-1})} &= \frac{x}{x-1}\\
\\
y &= \frac{x}{x -1 } && \text{(Equation 2)}\\
\\
\frac{d}{dx} (y) &= \frac{d}{dx} \left( \frac{x}{x-1} \right)\\
\\
\frac{dy}{dx} &= \frac{(x-1)\frac{d}{dx}(x)-(x) \frac{d}{dx}(x-1)}{(x-1)^2}\\
\\
\frac{dy}{dx} &= \frac{(x-1)(1)-(x)(1-0)}{(x-1)^2}\\
\\
\frac{dy}{dx} &= \frac{\cancel{x}-1-\cancel{x}}{(x-1)^2}\\
\\
\frac{dy}{dx} &= \frac{-1}{(x-1)^2} \qquad \text{ or } \qquad y' = \frac{-1}{(x-1)^2}
\end{aligned}
\end{equation}
$


c.) Substituting Equation 2 in Equation 1

$
\begin{equation}
\begin{aligned}
y' &= \frac{-y^2}{x^2} && \text{(Equation 1)}\\
\\
y &= \frac{x}{x-1} && \text{(Equation 2)}\\
\\
y' &= \frac{- \left( \frac{x}{x-1}\right)^2}{x^2}\\
\\
y' &= \frac{-\frac{x^2}{(x-1)^2}}{x^2}\\
\\
y' &= \frac{-\cancel{x^2}}{(\cancel{x^2})(x-1)^2}\\
\\
y' &= \frac{-1}{(x-1)^2}
\end{aligned}
\end{equation}
$


Results from part (a) and part(b) are equivalent.