Answer
$f'(t)=\frac{-1}{2(t+1)\sqrt{t+1}}$
The domain of $f'$ is $(-1,\infty)$.
Work Step by Step
$f(t)=\frac{1}{\sqrt{t+1}}$
Use the definition of a derivative.
$f'(t)=\lim\limits_{h \to 0}\frac{f(t+h)-f(t)}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{1}{\sqrt{t+h+1}}-\frac{1}{\sqrt{t+1}}}{h}$
$=\lim\limits_{h \to 0}\frac{\frac{\sqrt{t+1}-\sqrt{t+h+1}}{\sqrt{t+h+1}\sqrt{t+1}}}{h}$
$=\lim\limits_{h \to 0}{\frac{\sqrt{t+1}-\sqrt{t+h+1}}{h\sqrt{t+h+1}\sqrt{t+1}}}$
$=\lim\limits_{h \to 0}{\frac{\sqrt{t+1}-\sqrt{t+h+1}}{h\sqrt{t+h+1}\sqrt{t+1}}}\times
\frac{\sqrt{t+1}+\sqrt{t+h+1}}{\sqrt{t+1}+\sqrt{t+h+1}}$
$=\lim\limits_{h \to 0}\frac{t+1-(t+h+1)}{h\sqrt{t+h+1}\sqrt{t+1}(\sqrt{t+1}+\sqrt{t+h+1})}$
$=\lim\limits_{h \to 0}\frac{-h}{h\sqrt{t+h+1}\sqrt{t+1}(\sqrt{t+1}+\sqrt{t+h+1})}$
$=\lim\limits_{h \to 0}\frac{-1}{\sqrt{t+h+1}\sqrt{t+1}(\sqrt{t+1}+\sqrt{t+h+1})}$
$=\frac{-1}{\sqrt{t+0+1}\sqrt{t+1}(\sqrt{t+1}+\sqrt{t+0+1})}$
$=\frac{-1}{(t+1)\cdot 2\sqrt{t+1}}$
$=\frac{-1}{2(t+1)\sqrt{t+1}}$
Thus, $f'(t)=\frac{-1}{2(t+1)\sqrt{t+1}}$.
The domain of $f'$ is $(-1,\infty)$.
