# Devriative of $\frac {1} {\sqrt{x+1}}$ using first principle

I am stuck at the first step:

$$f(x)=\frac{1}{\sqrt{x+1}}$$ $$f'(x) = \lim_{h\to0} \frac{\frac{1}{\sqrt{h+x+1}}-\frac{1}{\sqrt{x+1}}}{h}$$

I tried multiplying by the conjugate but that didn't get me far, I also tried turning the function to $$(4+|x|)^{-1/2}$$ but that didn't work either

$$\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h+1}}-\frac{1}{\sqrt{x+1}}}{h}=\lim_{h\to0}\frac{\sqrt{x+1}-\sqrt{x+h+1}}{h\sqrt{(x+h+1)(x+1)}}=\lim_{h\to0}\frac{(x+1)-(x+h+1)}{h\sqrt{(x+h+1)(x+1)}(\sqrt{x+1}+\sqrt{x+h+1})}=\lim_{h\to0}\frac{-h}{h\sqrt{(x+h+1)(x+1)}(\sqrt{x+1}+\sqrt{x+h+1})}=\lim_{h\to0}\frac{-1}{\sqrt{(x+h+1)(x+1)}(\sqrt{x+1}+\sqrt{x+h+1})}=-\frac{1}{2(x+1)\sqrt{x+1}}=f'(x)$$
Hint: Suppose $x>0$ then $\frac{1}{\sqrt{1+|x+h|}}-\frac{1}{\sqrt{1+|x|}}=\frac{1}{\sqrt{1+x+h}}-\frac{1}{\sqrt{1+x}}$ so $\frac{\frac{1}{\sqrt{1+x+h}}-\frac{1}{\sqrt{1+x}}}{h}=\frac{\sqrt{1+x}-\sqrt{1+x+h}}{h\sqrt{(1+x+h)(1+x)}}=\frac{\sqrt{1+x}-\sqrt{1+x+h}}{h\sqrt{(1+x+h)(1+x)}}\times\frac{\Big(\sqrt{1+x}+\sqrt{1+x+h}\Big)}{\Big(\sqrt{1+x}+\sqrt{1+x+h}\Big)}=\frac{-h}{h\sqrt{(1+x)(1+x+h)}\big(\sqrt{1+x}+\sqrt{1+x+h}\big)}=\cdots$
The case $x<0$ is similar. (consider $|x|=-x$)
• Hello, I am very sorry I did a mistake in the body but the equation is supposed to be $$\frac{1}{\sqrt{x+1}}$$ – method Nov 28 '14 at 15:31