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use the defiition of derivitive to compute f'(3) f(x)=1+(1/(x-1))
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\[f(x)=1+\frac{1}{x-1}\] By the limit definition of the derivative, \[\begin{align*}f'(x)&=\lim_{h\to0}\frac{\left(1+\dfrac{1}{x+h-1}\right)-\left(1+\dfrac{1}{x-1}\right)}{h}\\\\ &=\lim_{h\to0}\frac{\dfrac{1}{x+h-1}-\dfrac{1}{x-1}}{h}\\\\ &=\lim_{h\to0}\frac{\dfrac{x-1}{(x-1)(x+h-1)}-\dfrac{x+h-1}{(x-1)(x+h-1)}}{h}\\\\ &=\lim_{h\to0}\frac{x-1-(x+h-1)}{h(x-1)(x+h-1)}\\\\ &=\lim_{h\to0}\frac{-h}{h(x-1)(x+h-1)}\\\\ &=\lim_{h\to0}\frac{-1}{(x-1)(x+h-1)}\\\\ &=-\frac{1}{(x-1)^2} \end{align*}\] Plug in \(x=3\) to find \(f'(3)\).
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