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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of the given function from first principle \[\] $ \Large\frac{x+1}{x-1}$

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Let $f(x) = \large\frac{x+1}{x-1}$. Accordingly, from the first principle.
$f'(x) = \lim\limits_{h \to 0}\large\frac{f(x+h)-f(x)}{h}$
$ = \lim\limits_{ h \to 0} \large\frac{ \bigg( \Large\frac{x+h+1}{x+h-1}-\Large\frac{x+1}{x-1}\bigg)}{h}$
$ = \lim\limits_{ h \to 0} \large\frac{1}{h} \bigg[ \large\frac{(x-1)(x+h+1)-(x+1)(x+h-1)}{(x-1)(x+h-1)}\bigg]$
$= \lim\limits_{ h \to 0} \large\frac{1}{h} \bigg[ \large\frac{(x^2+hx+1-x-h-1)-(x^2+hx-x+x+h-1)}{(x-1)(x+h-1)}\bigg]$
$= \lim\limits_{ h \to 0} \large\frac{1}{h} \bigg[ \large\frac{-2h}{(x-1)(x+h-1)}\bigg]$
$= \lim\limits_{ h \to 0} \large\frac{1}{h} \bigg[ \large\frac{-2}{(x-1)(x+h-1)}\bigg]$
$ = \large\frac{-2}{(x-1)(x-1)}$$ = \large\frac{-2}{(x-1)^2}$
answered Apr 7, 2014 by thanvigandhi_1

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