## Calculus 8th Edition

$f'(a) = \dfrac{2}{(1-a)\sqrt{1-a}}$
Given $f(x) = \dfrac{4}{\sqrt{1-x}}$ $f'(a) = \lim\limits_{x \to a} \dfrac{f(x) - f(a)}{x - a} = \lim\limits_{x \to a} \dfrac{ \dfrac{4}{\sqrt{1-x}} - \dfrac{4}{\sqrt{1-a}}}{x - a} = \lim\limits_{x \to a} \dfrac{ \dfrac{4\sqrt{1-a} - 4\sqrt{1-x}}{(\sqrt{1-x})(\sqrt{1-a})}}{x - a} = \lim\limits_{x \to a} \dfrac{4\sqrt{1-a} - 4\sqrt{1-x}}{(x - a)(\sqrt{1-x})(\sqrt{1-a})} = \lim\limits_{x \to a} \bigg[ \dfrac{4\sqrt{1-a} - 4\sqrt{1-x}}{(x - a)(\sqrt{1-x})(\sqrt{1-a})} \times \dfrac{4\sqrt{1-a} + 4\sqrt{1-x}}{4\sqrt{1-a} + 4\sqrt{1-x}}\bigg] = \lim\limits_{x \to a} \dfrac{(4\sqrt{1-a} - 4\sqrt{1-x})(4\sqrt{1-a} + 4\sqrt{1-x})}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{(4\sqrt{1-a})^2 - (4\sqrt{1-x})^2}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{16(1-a) - 16(1-x)}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{16-16a - 16 + 16x}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{16x - 16a}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{16(x - a)}{(x - a)(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \lim\limits_{x \to a} \dfrac{16}{(\sqrt{1-x})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-x})} = \dfrac{16}{(\sqrt{1-a})(\sqrt{1-a})(4\sqrt{1-a} + 4\sqrt{1-a})} = \dfrac{16}{(\sqrt{1-a})^2(8\sqrt{1-a})} = \dfrac{16}{8(1-a)\sqrt{1-a}} = \dfrac{2}{(1-a)\sqrt{1-a}} \longrightarrow f'(a) = \dfrac{2}{(1-a)\sqrt{1-a}}$