Answer
$$y''=\frac{1}{2}\Bigg[-\frac{1}{2\sqrt{x^3}(1-\sqrt x)^2}+\frac{1}{x(1-\sqrt x)^3}\Bigg]$$
Work Step by Step
$$y=(1-\sqrt x)^{-1}$$
- Find $y'$: $$y'=-(1-\sqrt x)^{-2}(1-\sqrt x)'=-(1-\sqrt x)^{-2}(1-x^{1/2})'$$ $$y'=-(1-\sqrt x)^{-2}(0-\frac{1}{2}x^{-1/2})=\frac{1}{2}x^{-1/2}(1-\sqrt x)^{-2}$$
- Find $y''$: $$y''=\frac{1}{2}(x^{-1/2}(1-\sqrt x)^{-2})'$$ $$y''=\frac{1}{2}\Bigg[(x^{-1/2})'(1-\sqrt x)^{-2}+x^{-1/2}\Big((1-\sqrt x)^{-2}\Big)'\Bigg]$$ $$y''=\frac{1}{2}\Bigg[-\frac{1}{2}x^{-3/2}(1-\sqrt x)^{-2}+x^{-1/2}\Big(-2(1-\sqrt x)^{-3}(1-\sqrt x)'\Big)\Bigg]$$ $$y''=\frac{1}{2}\Bigg[-\frac{1}{2\sqrt{x^3}(1-\sqrt x)^2}-2x^{-1/2}(1-\sqrt x)^{-3}\Big(-\frac{1}{2}x^{-1/2}\Big)\Bigg]$$ $$y''=\frac{1}{2}\Bigg[-\frac{1}{2\sqrt{x^3}(1-\sqrt x)^2}+x^{-1}(1-\sqrt x)^{-3}\Bigg]$$ $$y''=\frac{1}{2}\Bigg[-\frac{1}{2\sqrt{x^3}(1-\sqrt x)^2}+\frac{1}{x(1-\sqrt x)^3}\Bigg]$$
