Science:Math Exam Resources/Courses/MATH104/December 2015/Question 07/Solution 1

By definition,

$f'(a)=\lim _{h\to 0}{\frac {f(a+h)-f(a)}{h}}.$

Here $f({\color {Orange}x})={\tfrac {1}{2{\color {Orange}x}+1}}$ and $a=2.$ Hence we have

${\begin{aligned}f'(2)&=\lim _{h\to 0}{\frac {f({\color {OrangeRed}2+h})-f({\color {OliveGreen}2})}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {1}{2({\color {OrangeRed}2+h})+1}}-{\frac {1}{2({\color {OliveGreen}2})+1}}}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {1}{2h+5}}-{\frac {1}{5}}}{h}}\\&=\lim _{h\to 0}{\frac {\frac {5-(2h+5)}{5(2h+5)}}{h}}\\&=\lim _{h\to 0}\ {\frac {-2h}{5h(2h+5)}}\\&=\lim _{h\to {\color {Mulberry}0}}\ {\frac {-2}{5(2h+5)}}\\&={\frac {-2}{5(2\cdot {\color {Mulberry}0}+5)}}={\color {blue}-{\frac {2}{25}}}\,.\end{aligned}}$