Find the derivative of the functions from first principle $$\dfrac{x+1}{x-1}$$

Asked by Pragya Singh | 1 year ago |  62

##### Solution :-

Let f(x) $$\dfrac{x+1}{x-1}$$ Accordingly, from the first principle,

$$f'(x)=\lim\limits_{h \to 0}\dfrac{f(x+h)-f(x)}{h}$$

$$\lim\limits_{h \to 0}\dfrac{(\dfrac{x+h+1}{x+h-1}-\dfrac{x+1}{x-1})}{h}$$

$$\lim\limits_{h \to 0}\dfrac{1}{h}[\dfrac{-2h}{(x-1)(x+h-1)}]$$

$$\lim\limits_{h \to 0}[\dfrac{2}{(x-1)(x+h-1)}]$$

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

Answered by Abhisek | 1 year ago

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