Single Variable Calculus, Chapter 3, 3.6, Section 3.6, Problem 4

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: $\cos x + \sqrt{y} = 5$
$\displaystyle \frac{d}{dx} ( \cos x) + \frac{d}{dx} (\sqrt{y}) = \frac{d}{dx}(5)$
$\displaystyle \frac{d}{dx} (\cos x) + \frac{d}{dx} (y)^{\frac{1}{2}} = \frac{d}{dx}(5)$

$
\begin{equation}
\begin{aligned}
-\sin x + \frac{1}{2}(y)^{- \frac{1}{2}} \frac{d}{dx}(y) &= 0\\
\\
-\sin x + \frac{1}{2}(y)^{- \frac{1}{2}} \frac{dy}{dx} & = 0 \\
\\
-\sin x + \left( \frac{1}{2\sqrt{y}}\right) \frac{dy}{dx} & = 0\\
\\
\left( \frac{1}{2\sqrt{y}}\right) \frac{dy}{dx} & = \sin x\\
\\
\frac{dy}{dx} &= 2 \sqrt{y} \sin x && \text{(Equation 1)}
\end{aligned}
\end{equation}
$


b.) Solving for $y$

$
\begin{equation}
\begin{aligned}
\sqrt{y} &= 5 - \cos x \\
\\
(\sqrt{y})^2 &= (5 - \cos x)^2\\
\\
y & = (5- \cos x)^2 && \text{(Equation 2)}\\
\\
\frac{d}{dx} (y) & = \frac{d}{dx} ( 5- \cos x)^2\\
\\
\frac{dy}{dx} & = 2 (5- \cos x) \frac{d}{dx} (5- \cos x)\\
\\
\frac{dy}{dx} & = 2 (5 - \cos x ) \left[ 0 - (-\sin x) \right]\\
\\
\frac{dy}{dx} & = 2 (5 - \cos x ) ( \sin x ) \qquad \text{ or } \qquad y'=2(5-\cos x)(\sin x)
\end{aligned}
\end{equation}
$

c.) Substituting Equation 2 in Equation 1

$
\begin{equation}
\begin{aligned}
y' & = 2\sqrt{y} \sin x && \text{(Equation 1)}\\
\\
y &= (5-\cos x)^2 && \text{(Equation 2)}\\
\\
y' &= 2 \sqrt{(5-\cos x)^2} \sin x\\
\\
y' &= 2 (5-\cos x) \sin x
\end{aligned}
\end{equation}
$

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