Answer
$y''$ = $(\frac{6}{x^{3}})(1+\frac{1}{x})$$(1+\frac{2}{x})$
Work Step by Step
$y$ = $(1+\frac{1}{x})^{3}$
$y'$ = $\frac{d}{dx}$$(1+\frac{1}{x})^{3}$
$y'$ = $3(1+\frac{1}{x})^{2}$ $\frac{d}{dx}$$(1+\frac{1}{x})$
$y'$ = $3(1+\frac{1}{x})^{2}$$(-\frac{1}{x^{2}})$
$y''$ = $\frac{d}{dx}$[$3(1+\frac{1}{x})^{2}$$(-\frac{1}{x^{2}})$]
$y''$ = $3$[$(1+\frac{1}{x})^{2}$$\frac{d}{dx}$$(-\frac{1}{x^{2}})$ + $(-\frac{1}{x^{2}})$$\frac{d}{dx}$$(1+\frac{1}{x})^{2}$]
$y''$ = $3$[$(1+\frac{1}{x})^{2}$$(\frac{2}{x^{3}})$ + $(-\frac{1}{x^{2}})$$2(1+\frac{1}{x})$$\frac{d}{dx}$$(1+\frac{1}{x})$]
$y''$ = $3$[$(1+\frac{1}{x})^{2}$$(\frac{2}{x^{3}})$ + $(-\frac{1}{x^{2}})$$2(1+\frac{1}{x})$$(-\frac{1}{x^{2}})$]
$y''$ = $3$[$(1+\frac{1}{x})^{2}$$(\frac{2}{x^{3}})$ + $(\frac{2}{x^{4}})$$(1+\frac{1}{x})$]
$y''$ = $(\frac{6}{x^{3}})(1+\frac{1}{x})$[$(1+\frac{1}{x})$ + $(\frac{1}{x})$]
$y''$ = $(\frac{6}{x^{3}})(1+\frac{1}{x})$$(1+\frac{2}{x})$
