Single Variable Calculus, Chapter 3, 3.9, Section 3.9, Problem 14

a.) Determine the differential of $\displaystyle y = (t + \tan t)^5$

Using Differential Approximation

$dy = f'(t) dt$


$
\begin{equation}
\begin{aligned}

\frac{dy}{dt} =& \frac{d}{dt} (t + \tan t)^5
\\
\\
dy =& \left[ 5 (t + \tan t)^4 \frac{d}{dt} (t + \tan t) \right] dt
\\
\\
dy =& 5(t + \tan t)^4 (1 + \sec^2 t) dt


\end{aligned}
\end{equation}
$


b.) Determine the differential of $\displaystyle y = \sqrt{z + \frac{1}{2}}$

Using Differential Approximation

$dy = f'(z) dz$


$
\begin{equation}
\begin{aligned}

\frac{dy}{dz} =& \left(z + \frac{1}{2}\right)^{\frac{1}{2}}
\\
\\
dy =& \left[ \frac{1}{2} + \left( z + \frac{1}{z} \right) ^{\frac{-1}{2}} \frac{d}{dz} \left( z + \frac{1}{z} \right) \right] dz
\\
\\
dy =& \left[ \frac{1}{2} \left( z + \frac{1}{z} \right) ^{\frac{-1}{2}} \left( 1 + \left( \frac{-1}{z^2}\right) \right) \right] dz
\\
\\
dy =& \left[ \left( \frac{1}{\displaystyle 2 (z + \frac{1}{z})^{\frac{1}{2}}} \right) \left( 1 - \frac{1}{z^2} \right) \right] dz
\\
\\
dy =& \left[ \left( \frac{1}{\displaystyle 2 \left( \frac{z^2 + 1}{z} \right)^{\frac{1}{2}}} \right) \left( \frac{z^2 - 1}{z^2} \right) \right] dz
\\
\\
dy =& \left[ \frac{z^2 - 1}{2z^2 \displaystyle \frac{(z^2 + 1)^{\frac{1}{2}}}{(z)^{\frac{1}{2}}}} \right] dz
\\
\\
dy =& \left[ \frac{z^2 - 1}{2 (z)^{\frac{3}{2}} (z^2 + 1)^{\frac{1}{2}}} \right] dz
\\
\\
dy =& \left[ \frac{z^2 - 1}{2(z)(z)^{\frac{1}{2}} (z^2 + 1)^{\frac{1}{2}}} \right] dz
\\
\\
dy =& \frac{z^2 - 1}{2z [z (z^2 + 1)^{\frac{1}{2}}]^{\frac{1}{2}}} dz
\\
\\
dy =& \frac{z^2 - 1}{2z (z^3 + z)^{\frac{1}{2}}} dz
\\
\\
& \text{ or }
\\
\\
dy =& \frac{z^2 - 1}{2z \sqrt{z^3 + z}} dz

\end{aligned}
\end{equation}
$