Find the derivative of the function f(x) = fr | vedclass

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(A) Let $f(x) = \frac{x+1}{x-1}$. By the first principle of derivatives,$f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x+h)

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$\frac{-2}{(x-1)^{2}}$

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(A) Let $f(x) = \frac{x+1}{x-1}$. By the first principle of derivatives,$f^{\prime}(x) = \lim_{h 	o 0} \frac{f(x+h) - f(x)}{h}$.Substituting $f(x+h) = \frac{x+h+1}{x+h-1}$ and $f(x) = \frac{x+1}{x-1}$:$f^{\prime}(x) = \lim_{h 	o 0} \frac{1}{h} \left[ \frac{x+h+1}{x+h-1} - \frac{x+1}{x-1} ight]$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{(x-1)(x+h+1) - (x+1)(x+h-1)}{(x-1)(x+h-1)} ight]$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{(x^2 + xh + x - x - h - 1) - (x^2 + xh - x + x + h - 1)}{(x-1)(x+h-1)} ight]$$= \lim_{h 	o 0} \frac{1}{h} \left[ \frac{-2h}{(x-1)(x+h-1)} ight]$$= \lim_{h 	o 0} \frac{-2}{(x-1)(x+h-1)} = \frac{-2}{(x-1)^2}$.

Find the derivative of the function $f(x) = \frac{x+1}{x-1}$ using the first principle.

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