Single Variable Calculus, Chapter 3, 3.2, Section 3.2, Problem 25

Find the derivative of $\displaystyle G(t) = \frac{4t}{t + 1}$ using the definition and the domain of its derivative.

Using the definition of derivative


$
\begin{equation}
\begin{aligned}

\qquad G'(t) &= \lim_{h \to 0} \frac{g(t + h) - G(t)}{h}
&&
\\
\\
\qquad G'(t) &= \lim_{h \to 0} \frac{\displaystyle \frac{4(t + h)}{t + h + 1} - \frac{4t}{t + 1}}{h}
&& \text{Substitute $G(t + h)$ and $G(t)$}
\\
\\
\qquad G'(t) &= \lim_{h \to 0} \frac{(4t + 4h)(t + 1) - (4t)(t + h + 1)}{(h)(t + h + 1)(t + 1)}
&& \text{Get the LCD of the numerator}
\\
\\
\qquad G'(t) &= \lim_{h \to 0} \frac{\cancel{4t^2} + \cancel{4t} + \cancel{4th} + 4h - \cancel{4t^2} - \cancel{4th} - \cancel{4t}}{(h)(t + h + 1)(t + 1)}
&& \text{Expand and combine like terms}
\\
\\
\qquad G'(t) &= \lim_{h \to 0} \frac{4\cancel{h}}{\cancel{(h)} (t + h + 1)(t + 1)}
&& \text{Cancel out like terms}
\\
\\
\qquad G'(t) &= \lim_{h \to 0} \frac{4}{(t + h + 1)(t + 1)} = \frac{4}{(t + h + 1)(t + 1)} = \frac{4}{(t + 0 + 1)(t + 1)}
&& \text{Evaluate the limit}


\end{aligned}
\end{equation}
$


$\qquad \fbox{$G'(t) = \displaystyle \frac{4}{(t + 1)^2}$}$

Both functions are continuous on $0 > t + 1 > 0$.

$\displaystyle \begin{array}{ccc}
0 > & t + 1 & c \\
t > & -1 & \text{and } t > -1
\end{array} $

Therefore,

The domain of $G(t) = \displaystyle \frac{4t}{t + 1}$ is $(- \infty, -1)$

The domain of $G'(t) = \displaystyle \frac{4}{(t + 1)^2}$ is $(-\infty, -1) \bigcup (-1, \infty)$