Calculate the following derivative. (Use symbolic notation and fractions where needed.) \( \frac{d}{d x}\left(\frac{x+71}{x-71}\right)^{18}=\square \) help (fractions)

We start with \[ f(x)=\left(\frac{x+71}{x-71}\right)^{18}. \] Using the chain rule, we have \[ f'(x)=18\left(\frac{x+71}{x-71}\right)^{17}\cdot \frac{d}{dx}\left(\frac{x+71}{x-71}\right). \] Next, applying the quotient rule on \[ g(x)=\frac{x+71}{x-71}, \] we get \[ g'(x)=\frac{(x-71)\cdot\frac{d}{dx}(x+71)- (x+71)\cdot \frac{d}{dx}(x-71)}{(x-71)^2}. \] Since \[ \frac{d}{dx}(x+71)=1 \quad \text{and} \quad \frac{d}{dx}(x-71)=1, \] this becomes \[ g'(x)=\frac{(x-71)- (x+71)}{(x-71)^2}=\frac{-142}{(x-71)^2}. \] Substitute \( g'(x) \) into the derivative \( f'(x) \): \[ f'(x)=18\left(\frac{x+71}{x-71}\right)^{17}\cdot \left(-\frac{142}{(x-71)^2}\right). \] Combine the constants: \[ 18 \cdot (-142) = -2556, \] so \[ f'(x)=-2556\left(\frac{x+71}{x-71}\right)^{17}\frac{1}{(x-71)^2}. \] Expressing everything as a single fraction: \[ \left(\frac{x+71}{x-71}\right)^{17}\frac{1}{(x-71)^2}=\frac{(x+71)^{17}}{(x-71)^{17}(x-71)^2}=\frac{(x+71)^{17}}{(x-71)^{19}}. \] Thus, the derivative is \[ \boxed{\frac{d}{dx}\left(\frac{x+71}{x-71}\right)^{18} = -\frac{2556\,(x+71)^{17}}{(x-71)^{19}}}. \]