Science:Math Exam Resources/Courses/MATH100/December 2010/Question 06/Solution 2

Using the alternative definition,

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

Here we find a common denominator to simplify the fractions on top of the large fraction.

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

We then simplify the fraction, factoring out a 5 on top and then canceling the (x - t) term.

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

Finally we take the limit.

${\begin{aligned}f'(t)&={\frac {5}{(t+5)(t+5)}}\\f'(t)&={\frac {5}{(t+5)^{2}}}\\\end{aligned}}$

Which completes the question.