## Calculus (3rd Edition)

$$y'= -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$
Since $y=(1+\frac{1}{x})^3$, then by the chain rule $(f(g(x)))^{\prime}=f^{\prime}(g(x)) g^{\prime}(x)$, the derivative $y'$ is given by $$y'=3\left(1+\frac{1}{x}\right)^2\left(1+\frac{1}{x}\right)'\\=3\left(1+\frac{1}{x}\right)^2\left(-\frac{1}{x^2}\right) = -\frac{3}{x^2}\left(1+\frac{1}{x}\right)^2 .$$