Answer
$-\frac{1}{8}$
Work Step by Step
Step 1: Simplify the difference quotient $\frac{f(x)-f(a)}{x-a}$.
$\frac{f(x)-f(1)}{x-1}=\frac{\frac{1}{\sqrt{2x+2}}-\frac{1}{\sqrt{2\cdot 1+2}}}{x-1}=\frac{\frac{1}{\sqrt{2x+2}}-\frac{1}{2}}{x-1}=\frac{\frac{2-\sqrt{2x+2}}{2\sqrt{2x+2}}}{x-1}=\frac{2-\sqrt{2x+2}}{2(x-1)\sqrt{2x+2}}$
$=\frac{2-\sqrt{2x+2}}{2(x-1)\sqrt{2x+2}}\times \frac{2+\sqrt{2x+2}}{2+\sqrt{2x+2}}=\frac{4-(2x+2)}{2(x-1)\sqrt{2x+2}\cdot ({2+\sqrt{2x+2}})}=\frac{-2x+2}{2(x-1)\sqrt{2x+2}\cdot ({2+\sqrt{2x+2}})}$
$=\frac{-2(x-1)}{2(x-1)\sqrt{2x+2}\cdot ({2+\sqrt{2x+2}})}=\frac{-2}{2\sqrt{2x+2}\cdot ({2+\sqrt{2x+2}})}$
Step 2: Evaluate the limit $\lim\limits_{x \to a}\frac{f(x)-f(a)}{x-a}$.
$\lim\limits_{x \to 1}\frac{f(x)-f(1)}{x-1}=\lim\limits_{x \to 1}\frac{-2}{2\sqrt{2x+2}\cdot ({2+\sqrt{2x+2}})}=\frac{-2}{2\sqrt{2\cdot 1+2}\cdot ({2+\sqrt{2\cdot 1+2}})}$
$=\frac{-2}{2\cdot 2\cdot (2+2)}=-\frac{1}{8}$
Step 3. Conclude $f'(a)$.
$f'(1)=\lim\limits_{x \to 1}\frac{f(x)-f(1)}{x-1}=-\frac{1}{8}$
