MATH110 April 2016
• Q1 (a) • Q1 (b) • Q1 (c) • Q1 (d) • Q2 (a) • Q2 (b) • Q2 (c) • Q2 (d) • Q3 (a) • Q3 (b) • Q3 (c) • Q4 (a) • Q4 (b) • Q4 (c) • Q4 (d) • Q5 (a) • Q5 (b) • Q5 (c) • Q5 (d) • Q5 (e) • Q5 (f) • Q5 (g) • Q5 (h) • Q5 (i) • Q6 (a) • Q6 (b) • Q7 • Q8 (a) • Q8 (b) • Q8 (c) • Q8 (d) • Q9 (a) • Q9 (b) • Q10 •
Question 01 (d)

Express the derivative of $f(x)={\frac {1}{x^{2}+1}}$ at $x=2$ as a limit. If necessary, simplify your
answer so that it contains only the variable needed to evaluate the limit. You do not
need to evaluate the limit.

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 derivative of $f(x)$ is defined by $f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)f(x)}{h}}$

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Solution

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By hint, we have
$f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)f(x)}{h}}$,
so $f'(2)$ can be written as the limit
$f'(2)=\lim _{h\rightarrow 0}{\frac {f(2+h)f(2)}{h}}$.
To evaluate $f'(2)$ for $f(x)={\frac {1}{x^{2}+1}}$, we put the given $f$ into the formula;
$f'(2)=\lim _{h\rightarrow 0}{\frac {f(2+h)f(2)}{h}}=\lim _{h\rightarrow 0}{\frac {{\frac {1}{(2+h)^{2}+1}}{\frac {1}{2^{2}+1}}}{h}}=\lim _{h\rightarrow 0}{\frac {{\frac {1}{h^{2}+4h+5}}{\frac {1}{5}}}{h}}=\lim _{h\rightarrow 0}{\frac {1}{h}}\left({\frac {1}{h^{2}+4h+5}}{\frac {1}{5}}\right)$
Using the common denominator $5(h^{2}+4h+5)$, the difference of two fractions can be simplified as
${\frac {1}{h^{2}+4h+5}}{\frac {1}{5}}={\frac {5}{5(h^{2}+4h+5)}}{\frac {h^{2}+4h+5}{5(h^{2}+4h+5)}}={\frac {h^{2}+4h}{5(h^{2}+4h+5)}}={\frac {h(h+4)}{5(h^{2}+4h+5)}}$
Plugging this back, we have
$f'(2)=\lim _{h\rightarrow 0}{\frac {1}{h}}\left({\frac {1}{h^{2}+4h+5}}{\frac {1}{5}}\right)={\frac {1}{h}}\cdot \left({\frac {h(h+4)}{5(h^{2}+4h+5)}}\right)=\lim _{h\rightarrow 0}{\frac {(h+4)}{5(h^{2}+4h+5)}}=\lim _{h\rightarrow 0}{\frac {4}{5(0^{2}+4\cdot 0+5)}}={\frac {4}{25}}$
Answer $\color {blue}f'(2)=\lim _{h\rightarrow 0}{\frac {(h+4)}{5(h^{2}+4h+5)}}={\frac {4}{25}}$.
