Science:Math Exam Resources/Courses/MATH110/December 2011/Question 04 (a)/Solution 1

We simply input the function in the definition and compute the limit:

$\begin{array}{rl}{f}^{\prime }(x)& =\underset{h\to 0}{\lim }\frac{f(x+h)-f(x)}{h}\\ & =\underset{h\to 0}{\lim }\frac{\frac{1}{(x+h)+1}-\frac{1}{x+1}}{h}\\ & =\underset{h\to 0}{\lim }\frac{1}{h}(\frac{1}{(x+h+1)}-\frac{1}{(x+1)})\\ & =\underset{h\to 0}{\lim }\frac{1}{h}(\frac{x+1}{(x+h+1)(x+1)}-\frac{x+h+1}{(x+h+1)(x+1)})\\ & =\underset{h\to 0}{\lim }\frac{1}{h}\cdot \frac{(x+1)-(x+h+1)}{(x+h+1)(x+1)}\\ & =\underset{h\to 0}{\lim }\frac{1}{h}\cdot \frac{-h}{(x+h+1)(x+1)}\\ & =\underset{h\to 0}{\lim }-\frac{1}{(x+h+1)(x+1)}\\ & =-\frac{1}{(x+1)(x+1)}\\ & =-\frac{1}{(x+1{)}^2}\end{array}$