Here we first calculate the derivative for any value x {\displaystyle x} and only then plug in the value a = 1 {\displaystyle a=1} . That is, we first calculate
We use this to evaluate the derivative of the given function:
f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 ( x + h ) 2 ( x + h ) 2 + 1 − x 2 x 2 + 1 h = lim h → 0 ( x + h ) 2 ( x 2 + 1 ) − x 2 ( ( x + h ) 2 + 1 ) h ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) = lim h → 0 ( 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 → 0 ( x + h ) 2 − x 2 h ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) = lim h → 0 x 2 + 2 x h + h 2 − x 2 h ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) = lim h → 0 h ( 2 x + h ) h ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) = lim h → 0 ( 2 x + h ) ( ( x + h ) 2 + 1 ) ( x 2 + 1 ) = 2 x ( x 2 + 1 ) 2 {\displaystyle {\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 ) = 2 ( 1 ) ( 1 2 + 1 ) 2 = 1 2 {\displaystyle f'(1)={\frac {2(1)}{(1^{2}+1)^{2}}}={\frac {1}{2}}} .
and only then plug in the value 
. That is, we first calculate
.
