Answer
$$y'=\frac{20x(1-x)+x^2+1}{2(x^2+1)(1-x)}$$
Work Step by Step
$$y=\ln\Big(\frac{(x^2+1)^5}{\sqrt{1-x}}\Big)$$
The derivative of $y$: $$y'=\frac{1}{\Big(\frac{(x^2+1)^5}{\sqrt{1-x}}\Big)}\times\Big(\frac{(x^2+1)^5}{\sqrt{1-x}}\Big)'$$
$$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{\Big((x^2+1)^5\Big)'\sqrt{1-x}-(x^2+1)^5(\sqrt{1-x})'}{1-x}$$
We have $$\Big((x^2+1)^5\Big)'=5(x^2+1)^4(x^2+1)'=5(x^2+1)^4(2x)=10x(x^2+1)^4$$ $$(\sqrt{1-x})'=\frac{1}{2\sqrt{1-x}}(1-x)'=\frac{1}{2\sqrt{1-x}}(-1)=-\frac{1}{2\sqrt{1-x}}$$
Therefore, $$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{10x(x^2+1)^4\sqrt{1-x}-\Big(-\frac{(x^2+1)^5}{2\sqrt{1-x}}\Big)}{1-x}$$
$$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{10x(x^2+1)^4\sqrt{1-x}+\frac{(x^2+1)^5}{2\sqrt{1-x}}}{1-x}$$
$$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2\sqrt{1-x}}}{1-x}$$
$$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2(1-x)\sqrt{1-x}}$$
$$y'=\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2(x^2+1)^5(1-x)}$$
$$y'=\frac{20x(1-x)+x^2+1}{2(x^2+1)(1-x)}$$
