## Calculus: Early Transcendentals 8th Edition

$$g'(x)=\frac{-1}{2\sqrt{9-x}}$$ The domain of $g(x)$ is $(-\infty,9]$ The domain of $g'(x)$ is $(-\infty,9)$
$$g(x)=\sqrt{9-x}$$ According to definition, $$g'(x)=\lim\limits_{h\to0}\frac{g(x+h)-g(x)}{h}$$ $$g'(x)=\lim\limits_{h\to0}\frac{\sqrt{9-x-h}-\sqrt{9-x}}{h}$$ Multiply both numerator and denominator by $(\sqrt{9-x-h}+\sqrt{9-x})$, the numerator would become $$(\sqrt{9-x-h}-\sqrt{9-x})(\sqrt{9-x-h}+\sqrt{9-x})$$ $$=(9-x-h)-(9-x)$$ $$=9-x-h-9+x$$ $$=-h$$ Therefore, $$g'(x)=\lim\limits_{h\to0}\frac{-h}{h(\sqrt{9-x-h}+\sqrt{9-x})}$$ $$g'(x)=\lim\limits_{h\to0}\frac{-1}{(\sqrt{9-x-h}+\sqrt{9-x})}$$ $$g'(x)=\frac{-1}{\sqrt{9-x}+\sqrt{9-x}}$$ $$g'(x)=\frac{-1}{2\sqrt{9-x}}$$ *Domain of $g(x)$: $(9-x)$ must be $\ge0$. So, $x\le9$ Therefore, the domain of $g(x)$ is $(-\infty,9]$ *Domain of $g'(x)$: As before, $(9-x)$ must be $\ge0$. So, $x\le9$ Also, $2\sqrt{9-x}\ne0$. Therefore, $9-x\ne0$, so $x\ne9$ After all, the domain of $g'(x)$ is $(-\infty,9)$