Calculus and Its Applications, Chapter 1, 1.6, Section 1.6, Problem 20

Take the derivative of $\displaystyle y = \frac{t^2 - 25}{t - 5}$: first, use the Quotient Rule;
then, by dividing the expression before differentiating. Compare your results as a check.
By using Quotient Rule,

$
\begin{equation}
\begin{aligned}
y' &= \frac{(t - 5) \cdot \frac{d}{dt} (t^2 - 25) - (t^2 - 25) \cdot \frac{d}{dt} (t - 5)}{(t - 5)^2}\\
\\
y' &= \frac{(t - 5)(2t) - (t^2 - 25)(1)}{(t - 5)^2}\\
\\
y' &= \frac{2t^2 - 10t - t^2 + 25}{(t - 5)^2}\\
\\
&= \frac{t^2 - 10t + 25}{t^2 - 10t + 25} \\
\\
&= 1
\end{aligned}
\end{equation}
$



By dividing the expression first and recalling the factor difference of square we obtaion

$
\begin{equation}
\begin{aligned}
y &= \frac{t^2 - 25}{t - 5} = \frac{(t + 5)(t - 5)}{(t - 5)} = t + 5 \\
\\
y' &= \frac{d}{dt} ( t + 5) = 1
\end{aligned}
\end{equation}
$


Both results agree.