Question

use the defiition of derivitive to compute f'(3)
f(x)=1+(1/(x-1))

Answer 1

pleasee hurry!!

Answer 2

\[f(x)=1+\frac{1}{x-1}\]
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 \(f'(3)\).