[LIMIT] Use limit to calculate f'(x), f(x)=1/sqt x

[Dreams81]

Use Limit to calculate f'(x)

\(\displaystyle \L\,f(x) = \frac{1}{(sqrt x)}\)

Can some one show me step by step to do this, i'm completely lost.

Thanks in advanced

 

[tkhunny]

You should have a formula. Something like this:

\(\displaystyle \lim_{h->0}{\frac{f(x+h) - f(x)}{h}}\)

Have you seen it? That would be where to start.

 

[soroban]

Hello, Dreams81!

This is a tricky one . . .

I assume you know the formula: \(\displaystyle \:\:f'(x) \:=\:\lim_{h\to0}\frac{f(x\,+\,h)\,-\,f(x)}{h}\)

\(\displaystyle \text{Use limit to calculate }f'(x):\;f(x) \:= \:\L\frac{1}{sqrt x}\)

\(\displaystyle f(x\,+\,h)\,-\,f(x)\;=\;\L\frac{1}{\sqrt{x\,+\,h}} \,- \,\frac{1}{\sqrt{x}} \;= \;\frac{\sqrt{x}\,-\,\sqrt{x\,+\,h}}{\sqrt{x}\cdot\sqrt{x\,+\,h}}\)

Rationalize: \(\displaystyle \L\:\frac{\sqrt{x}\,-\,\sqrt{x\,+\,h}}{\sqrt{x}\cdot\sqrt{x\,+\,h}}\cdot\frac{\sqrt{x}\,+\,\sqrt{x\.+\,h}}{\sqrt{x}\,+\,\sqrt{x\.+\.h}} \;=\;\frac{x\,-\,(x\,+\,h)}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)} \;= \;\frac{-h}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h)}\)

Divide by \(\displaystyle h: \L\;\;\frac{f(x\,+\,h)\,-\,f(x)}{h}\;=\;\frac{1}{h}\cdot\frac{-h}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)} \;= \;\frac{-1}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)}\)

Take the limit: \(\displaystyle \;f'(x) \;= \;\lim_{h\to0}\L\left[\frac{-1}{\sqrt{x}\cdot\sqrt{x\,+\,h}\left(\sqrt{x}\,+\,\sqrt{x\,+\,h}\right)}\right] \;= \;\frac{-1}{\sqrt{x}\cdot\sqrt{x}\left(\sqrt{x}\,+\,\sqrt{x}\right)}\)

Therefore: \(\displaystyle \;f'(x)\;=\;\L\frac{-1}{x\cdot2\sqrt{x}}\)\(\displaystyle \;= \;\L-\frac{1}{2x^{3/2}}\)