Medium
Find the derivative of the following function: $\frac{1+\frac{1}{x}}{1-\frac{1}{x}}$
Let $f(x) = \frac{1+\frac{1}{x}}{1-\frac{1}{x}} = \frac{\frac{x+1}{x}}{\frac{x-1}{x}} = \frac{x+1}{x-1}$,where $x \neq 0$ and $x \neq 1$.
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$f'(x) = \frac{(x-1) \frac{d}{dx}(x+1) - (x+1) \frac{d}{dx}(x-1)}{(x-1)^2}$
$f'(x) = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$
$f'(x) = \frac{x - 1 - x - 1}{(x-1)^2}$
$f'(x) = \frac{-2}{(x-1)^2}$
Using the quotient rule,$\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$:
$f'(x) = \frac{(x-1) \frac{d}{dx}(x+1) - (x+1) \frac{d}{dx}(x-1)}{(x-1)^2}$
$f'(x) = \frac{(x-1)(1) - (x+1)(1)}{(x-1)^2}$
$f'(x) = \frac{x - 1 - x - 1}{(x-1)^2}$
$f'(x) = \frac{-2}{(x-1)^2}$
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