## University Calculus: Early Transcendentals (3rd Edition)

$$y''=\frac{6}{x^3}\Big(1+\frac{1}{x}\Big)\Big(1+\frac{2}{x}\Big)$$
$$y=\Big(1+\frac{1}{x}\Big)^3$$ - Find $y'$: $$y'=3\Big(1+\frac{1}{x}\Big)^2\Big(1+\frac{1}{x}\Big)'=3\Big(1+\frac{1}{x}\Big)^2\Big(0+\frac{(1)'x-1(x)'}{x^2}\Big)$$ $$y'=3\Big(1+\frac{1}{x}\Big)^2\Big(\frac{0\times x-1\times1}{x^2}\Big)=3\Big(1+\frac{1}{x}\Big)^2\Big(-\frac{1}{x^2}\Big)$$ $$y'=-\frac{3}{x^2}\Big(1+\frac{1}{x}\Big)^2$$ - Find $y''$: $$y''=\Big(-\frac{3}{x^2}\Big)'\Big(1+\frac{1}{x}\Big)^2+\Big(-\frac{3}{x^2}\Big)\Bigg[\Big(1+\frac{1}{x}\Big)^2\Bigg]'$$ $$y''=-\Big(\frac{(3)'x^2-3(x^2)'}{x^4}\Big)\Big(1+\frac{1}{x}\Big)^2-\frac{3}{x^2}\Bigg[2\Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{x}\Big)'\Bigg]$$ $$y''=-\Big(\frac{0x^2-6x}{x^4}\Big)\Big(1+\frac{1}{x}\Big)^2-\frac{3}{x^2}\Bigg[2\Big(1+\frac{1}{x}\Big)\Big(-\frac{1}{x^2}\Big)\Bigg]$$ $$y''=\frac{6}{x^3}\Big(1+\frac{1}{x}\Big)^2+\frac{6}{x^4}\Big(1+\frac{1}{x}\Big)$$ $$y''=\frac{6}{x^3}\Big(1+\frac{1}{x}\Big)\Big(1+\frac{1}{x}+\frac{1}{x}\Big)$$ $$y''=\frac{6}{x^3}\Big(1+\frac{1}{x}\Big)\Big(1+\frac{2}{x}\Big)$$