MATH104 December 2014
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Question 02

Let $f(x)={\frac {x^{2}}{x^{2}+1}}$. Use the definition of derivative to find $f'(1)$. No marks will be given for the use of any differentiation rules.

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

There are two equivalent limit definitions of the derivative at the point $a$:
 $f'(a)=\lim _{h\to 0}{\frac {f(a+h)f(a)}{h}}$
and
 $f'(a)=\lim _{x\to a}{\frac {f(x)f(a)}{xa}}$
given that these limits exist. You can use either definition. See Solution 1 for the former approach, and Solution 2 for the latter.

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Solution 1

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Here we first calculate the derivative for any value $x$ and only then plug in the value $a=1$. That is, we first calculate
 $f'(x)=\lim _{h\rightarrow 0}{\frac {f(x+h)f(x)}{h}}$
We use this to evaluate the derivative of the given function:
${\begin{aligned}f'(x)&=\lim _{h\rightarrow 0}{\frac {f(x+h)f(x)}{h}}\\&=\lim _{h\rightarrow 0}{\frac {{\frac {(x+h)^{2}}{(x+h)^{2}+1}}{\frac {x^{2}}{x^{2}+1}}}{h}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}(x^{2}+1)x^{2}((x+h)^{2}+1)}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}x^{2}+(x+h)^{2}x^{2}(x+h)^{2}x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(x+h)^{2}x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {x^{2}+2xh+h^{2}x^{2}}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {h(2x+h)}{h((x+h)^{2}+1)(x^{2}+1)}}\\&=\lim _{h\rightarrow 0}{\frac {(2x+h)}{((x+h)^{2}+1)(x^{2}+1)}}\\&={\frac {2x}{(x^{2}+1)^{2}}}\end{aligned}}$
Lastly, plugging in we obtain
$f'(1)={\frac {2(1)}{(1^{2}+1)^{2}}}={\frac {1}{2}}$.

Solution 2

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In this solution we use the formula
$f'(a)=\lim _{x\to a}{\frac {f(x)f(a)}{xa}}$
We could calculate $f'(a)$ for any value of $a$ and then plug in $a=1$, as it was done in Solution 1, but to show an alternative we plug in $a=1$ directly. Following the above definition of the derivative and the definition of the function we obtain:
${\begin{aligned}f'(1)&=\lim _{x\to 1}{\frac {f(x)f(1)}{x1}}\\&=\lim _{x\to 1}{\frac {{\frac {x^{2}}{x^{2}+1}}{\frac {1^{2}}{1^{2}+1}}}{x1}}\\&=\lim _{x\to 1}{\frac {{\frac {2x^{2}}{2(x^{2}+1)}}{\frac {x^{2}+1}{2(x^{2}+1)}}}{x1}}\\&=\lim _{x\to 1}{\frac {2x^{2}(x^{2}+1)}{2(x1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x^{2}1}{2(x1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {(x+1)(x1)}{2(x1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x+1}{2(x^{2}+1)}}\\&={\frac {1+1}{2(1^{2}+1)}}\\&={\frac {1}{2}}\end{aligned}}$

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