Science:Math Exam Resources/Courses/MATH100/December 2010/Question 06/Solution 1

By the definition of a derivative, we have that

${\begin{aligned}f'(t)&=\lim _{h\to 0}{\frac {f(t+h)-f(t)}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {t+h}{t+h+5}}-{\frac {t}{t+5}}}{h}}\end{aligned}}$

The key step to simplifying this expression is here: we find a common denominator for the fractions on top and combine them.

${\begin{aligned}&=\lim _{h\to 0}{\frac {{\frac {(t+h)(t+5)}{(t+h+5)(t+5)}}-{\frac {t(t+h+5)}{(t+h+5)(t+5)}}}{h}}\\&=\lim _{h\to 0}{\frac {\frac {t^{2}+th+5t+5h-t^{2}-th-5t}{(t+h+5)(t+5)}}{h}}\end{aligned}}$

Now we simplify the expression into a single fraction and cancel the h on top and bottom.

${\begin{aligned}&=\lim _{h\to 0}{\frac {5h}{h(t+h+5)(t+5)}}\\&=\lim _{h\to 0}{\frac {5}{(t+h+5)(t+5)}}\end{aligned}}$

Finally taking the limit gives:

${\begin{aligned}&={\frac {5}{(t+5)(t+5)}}\\&={\frac {5}{(t+5)^{2}}}\end{aligned}}$