I understand the quotient rule (and would be much easier), but I have to use the definition; because of that I've been stuck trying to solve this, and need help solving this using the definition:
$$ \frac{\mathrm{d}}{\mathrm{d}x} \frac{1}{\sqrt{x+1}} $$
I feel like I've missed something really simple, but I can't point to it.
On
Just as an alternative solution that uses the alternative (and equivalent) definition of a derivative,
$$f'(a) = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} = \lim_{x \to a}\frac{\frac{1}{\sqrt{x+1}} -\frac{1}{\sqrt{a+1}}}{x-a} $$$$= \lim_{x \to a}\frac{\sqrt{a+1}-\sqrt{x+1}}{(x-a)\sqrt{(x+1)(a+1)}} \cdot \frac{\sqrt{a+1}+\sqrt{x+1}}{\sqrt{a+1}+\sqrt{x+1}}$$ $$=\lim_{x \to a}\frac{(a+1)-(x+1)}{(x-a)\sqrt{(x+1)(a+1)}(\sqrt{a+1}+\sqrt{x+1})} $$$$= \lim_{x \to a}\frac{-(x-a)}{(x-a)\sqrt{(x+1)(a+1)}(\sqrt{a+1}+\sqrt{x+1})}$$ $$=\lim_{x \to a}\frac{-1}{\sqrt{(x+1)(a+1)}(\sqrt{a+1}+\sqrt{x+1})} $$$$= \frac{-1}{\sqrt{(a+1)(a+1)}(\sqrt{a+1}+\sqrt{a+1})} = \frac{-1}{2(a+1)\sqrt{a+1}}$$
$$=\lim_{h\to 0}{\frac{\frac{1}{\sqrt{x+h+1}}-\frac{1}{\sqrt{x+1}}}{h}}$$ $$=\lim_{h\to 0}{\frac{\sqrt{x+1}-\sqrt{x+h+1}}{h\sqrt{x+1}\sqrt{x+h+1}}}$$ $$=\big(\lim_{h\to 0}{\frac{1}{\sqrt{x+1}\sqrt{x+h+1}}}\big)\cdot\big(\lim_{h\to 0}{\frac{\sqrt{x+1}-\sqrt{x+h+1}}{h}}\big)$$ $$=\frac{1}{x+1}\cdot\big(\lim_{h\to 0}{\frac{(\sqrt{x+1}+\sqrt{x+h+1})(\sqrt{x+1}-\sqrt{x+h+1})}{(\sqrt{x+1}+\sqrt{x+h+1})h}}\big)$$ $$=\frac{1}{x+1}\cdot\big(\lim_{h\to 0}{\frac{-1}{(\sqrt{x+1}+\sqrt{x+h+1})}}\big)$$ $$-\frac{1}{2}(x+1)^{-\frac{3}{2}}$$