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.