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

$$y'=\frac{20x(1-x)+x^2+1}{2(x^2+1)(1-x)}$$
$$y=\ln\Big(\frac{(x^2+1)^5}{\sqrt{1-x}}\Big)$$ 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, $$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{10x(x^2+1)^4\sqrt{1-x}-\Big(-\frac{(x^2+1)^5}{2\sqrt{1-x}}\Big)}{1-x}$$ $$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{10x(x^2+1)^4\sqrt{1-x}+\frac{(x^2+1)^5}{2\sqrt{1-x}}}{1-x}$$ $$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2\sqrt{1-x}}}{1-x}$$ $$y'=\frac{\sqrt{1-x}}{(x^2+1)^5}\times\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2(1-x)\sqrt{1-x}}$$ $$y'=\frac{20x(x^2+1)^4(1-x)+(x^2+1)^5}{2(x^2+1)^5(1-x)}$$ $$y'=\frac{20x(1-x)+x^2+1}{2(x^2+1)(1-x)}$$