Science:Math Exam Resources/Courses/MATH104/December 2014/Question 02/Solution 2

In this solution we use the formula

$f'(a)=\lim _{x\to a}{\frac {f(x)-f(a)}{x-a}}$

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:

${\begin{aligned}f'(1)&=\lim _{x\to 1}{\frac {f(x)-f(1)}{x-1}}\\&=\lim _{x\to 1}{\frac {{\frac {x^{2}}{x^{2}+1}}-{\frac {1^{2}}{1^{2}+1}}}{x-1}}\\&=\lim _{x\to 1}{\frac {{\frac {2x^{2}}{2(x^{2}+1)}}-{\frac {x^{2}+1}{2(x^{2}+1)}}}{x-1}}\\&=\lim _{x\to 1}{\frac {2x^{2}-(x^{2}+1)}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x^{2}-1}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {(x+1)(x-1)}{2(x-1)(x^{2}+1)}}\\&=\lim _{x\to 1}{\frac {x+1}{2(x^{2}+1)}}\\&={\frac {1+1}{2(1^{2}+1)}}\\&={\frac {1}{2}}\end{aligned}}$