Question

Find the derivative of f(x)=x+1/

from the first principle​

Answer 1

Correct question :

Find the deravation of $\sf \dfrac{x + 1}{x - 1}$ from the first principle.

Answer

$\sf f(x) = \dfrac{x + 1}{x - 1}$

we know that,

$\boxed{\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{f(x +h) - f(x)}{h} }$

$\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{ \dfrac{(x + h + 1)}{(x + h - 1)} - \bigg( \dfrac{x + 1}{x - 1} \bigg) }{h}$

$\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{(x + h + 1)(x - 1) - (x + 1)(x + h - 1)}{h(x + h - 1)(x - 1)}$

$\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{ {x}^{2} + hx + x - x - h - 1 - hx + x - x - h + 1 }{h(x + h - 1)(x - 1)}$

$\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{ - 2h}{h(x + h - 1)(x - 1)}$

$\displaystyle \sf f(x) = \lim_{h \to 0} \dfrac{ - 2}{(x + h - 1)(x - 1)}$

$\displaystyle \sf f(x) = \dfrac{ - 2}{(x - 1)^{2} }$