## Calculus: Early Transcendentals 8th Edition

$$f'(a)=\frac{2}{(1-a)^{3/2}}$$
$$f(x)=\frac{4}{\sqrt{1-x}}$$ That means $$f(a)=\frac{4}{\sqrt{1-a}}$$ The derivative of $f(x)$ at number $a$ can be found as follows: $$f'(a)=\lim\limits_{x\to a}\frac{f(x)-f(a)}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{\frac{4}{\sqrt{1-x}}-\frac{4}{\sqrt{1-a}}}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{\frac{4\sqrt{1-a}-4\sqrt{1-x}}{\sqrt{(1-x)}\sqrt{(1-a)}}}{x-a}$$ $$f'(a)=\lim\limits_{x\to a}\frac{4\sqrt{1-a}-4\sqrt{1-x}}{(x-a)\sqrt{(1-x)}\sqrt{(1-a)}}$$ Multiply both numerator and denominator by $(4\sqrt{1-a}+4\sqrt{1-x})$, the numerator would become $$(4\sqrt{1-a}-4\sqrt{1-x})(4\sqrt{1-a}+4\sqrt{1-x})$$$$=16(1-a)-16(1-x)$$$$=16-16a-16+16x$$$$=16x-16a$$ Therefore, $$f'(a)=\lim\limits_{x\to a}\frac{16x-16a}{(x-a)\sqrt{(1-x)}\sqrt{(1-a)}(4\sqrt{1-a}+4\sqrt{1-x})}$$ $$f'(a)=\lim\limits_{x\to a}\frac{16}{\sqrt{(1-x)}\sqrt{(1-a)}(4\sqrt{1-a}+4\sqrt{1-x})}$$ $$f'(a)=\frac{16}{\sqrt{(1-a)}\sqrt{(1-a)}(4\sqrt{1-a}+4\sqrt{1-a})}$$ $$f'(a)=\frac{16}{\sqrt{(1-a)}^2(8\sqrt{1-a})}$$ $$f'(a)=\frac{2}{(1-a)\sqrt{1-a}}$$ $$f'(a)=\frac{2}{(1-a)^{3/2}}$$