By the limit definition of the derivative of f {\displaystyle f} at x {\displaystyle x} , f ′ ( x ) = lim h → 0 f ( x + h ) − f ( x ) h = lim h → 0 x + h 2 ( x + h ) + 1 − x 2 x + 1 h = lim h → 0 ( x + h ) ( 2 x + 1 ) − x ( 2 ( x + h ) + 1 ) h ( 2 ( x + h ) + 1 ) ( 2 x + 1 ) = lim h → 0 ( 2 x 2 + x + 2 x h + h ) − ( 2 x 2 + 2 x h + x ) h ( 2 ( x + h ) + 1 ) ( 2 x + 1 ) = lim h → 0 h h ( 2 ( x + h ) + 1 ) ( 2 x + 1 ) = lim h → 0 1 ( 2 ( x + h ) + 1 ) ( 2 x + 1 ) = 1 ( 2 ( x + 0 ) + 1 ) ( 2 x + 1 ) = 1 ( 2 x + 1 ) 2 {\displaystyle {\begin{aligned}f'(x)&=\lim _{h\to 0}{\frac {f(x+h)-f(x)}{h}}\\&=\lim _{h\to 0}{\frac {{\frac {x+h}{2(x+h)+1}}-{\frac {x}{2x+1}}}{h}}\\&=\lim _{h\to 0}{\frac {(x+h)(2x+1)-x(2(x+h)+1)}{h(2(x+h)+1)(2x+1)}}\\&=\lim _{h\to 0}{\frac {(2x^{2}+x+2xh+h)-(2x^{2}+2xh+x)}{h(2(x+h)+1)(2x+1)}}\\&=\lim _{h\to 0}{\frac {h}{h(2(x+h)+1)(2x+1)}}\\&=\lim _{h\to 0}{\frac {1}{(2(x+h)+1)(2x+1)}}\\&={\frac {1}{(2(x+0)+1)(2x+1)}}\\&={\color {blue}{\frac {1}{(2x+1)^{2}}}}\end{aligned}}}
Use of the quotient rule confirms that this is indeed f ′ ( x ) {\displaystyle \displaystyle f'(x)} : f ′ ( x ) = 1 ⋅ ( 2 x + 1 ) − x ⋅ 2 ( 2 x + 1 ) 2 = 1 ( 2 x + 1 ) 2 {\displaystyle {\begin{aligned}f'(x)&={\frac {1\cdot (2x+1)-x\cdot 2}{(2x+1)^{2}}}\\&={\color {OliveGreen}{\frac {1}{(2x+1)^{2}}}}\end{aligned}}}
