MATH110 December 2011
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Question 04 (a)

Let
 $f(x)={\frac {1}{x+1}}$
Find ƒ'(x) using the limit definition of the derivative.
No marks will be given for using other methods in part (a).

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint

The definition of the derivative is
 $f'(x)=\lim _{h\to 0}{\frac {f(x+h)f(x)}{h}}$

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Solution

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We simply input the function in the definition and compute the limit:
 ${\begin{aligned}f'(x)&=\lim _{h\to 0}{\frac {f(x+h)f(x)}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {1}{(x+h)+1}}{\frac {1}{x+1}}}{h}}\\&=\lim _{h\to 0}{\frac {1}{h}}\left({\frac {1}{(x+h+1)}}{\frac {1}{(x+1)}}\right)\\&=\lim _{h\to 0}{\frac {1}{h}}\left({\frac {x+1}{(x+h+1)(x+1)}}{\frac {x+h+1}{(x+h+1)(x+1)}}\right)\\&=\lim _{h\to 0}{\frac {1}{h}}\cdot {\frac {(x+1)(x+h+1)}{(x+h+1)(x+1)}}\\&=\lim _{h\to 0}{\frac {1}{h}}\cdot {\frac {h}{(x+h+1)(x+1)}}\\&=\lim _{h\to 0}{\frac {1}{(x+h+1)(x+1)}}\\&={\frac {1}{(x+1)(x+1)}}\\&={\frac {1}{(x+1)^{2}}}\end{aligned}}$

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