Derivate

**Petrus** · February 20th 2013, 04:56 AM

Hello!I got big problem derivate $\frac{1}{sqrt(x)}$
I basicly know how to derivate this without the special derivate formula i wanna train on. I wanna be able to derivate with lim h->0 $\frac{f(a+h)-f(a)}{h}$

**Prove It** · February 20th 2013, 05:10 AM

$\displaystyle \begin{align*} f(x) &= \frac{1}{\sqrt{x}} \\ \\ f'(x) &= \lim_{h \to 0}\frac{f(x + h) - f(x)}{h} \\ &= \lim_{h \to 0}\frac{\frac{1}{\sqrt{x + h}} - \frac{1}{\sqrt{x}}}{h} \\ &= \lim_{h \to 0}\frac{\frac{\sqrt{x} - \sqrt{x + h}}{\sqrt{x + h}\sqrt{x}}}{h} \\ &= \lim_{h \to 0}\frac{\sqrt{x} - \sqrt{x + h}}{h\sqrt{x + h}\sqrt{x}} \\ &= \lim_{h \to 0}\frac{ \left( \sqrt{x} - \sqrt{x + h} \right) \left( \sqrt{x} + \sqrt{x + h} \right) }{h \sqrt{x+h} \sqrt{x} \left( \sqrt{x} + \sqrt{x + h} \right) } \\ &= \lim_{h \to 0} \frac{x - \left( x + h \right) }{ h \sqrt{x + h} \sqrt{x} \left( \sqrt{x} + \sqrt{x + h} \right) } \\ &= \lim_{h \to 0} \frac{-h}{h\sqrt{x + h}\sqrt{x} \left( \sqrt{x} + \sqrt{x + h} \right) } \\ &= \lim_{h \to 0}\frac{-1}{\sqrt{x + h} \sqrt{x} \left( \sqrt{x} + \sqrt{x + h} \right) } \\ &= \frac{-1}{\sqrt{x + 0} \sqrt{x} \left( \sqrt{x} + \sqrt{x + 0} \right) } \end{align*}$

$\displaystyle \begin{align*} &= \frac{-1}{\sqrt{x} \sqrt{x} \left( \sqrt{x} + \sqrt{x} \right) } \\ &= \frac{-1}{2x\sqrt{x}} \end{align*}$

**MathJack** · February 20th 2013, 05:11 AM

1/(sqrt(x) + h) - 1/(sqrt(x))
multiply top and bottom by conjugate
in answer add numerator
then let h = 0

**Petrus** · February 20th 2013, 05:17 AM

I see what i did wrong on My progress!! Ty alot

Well i had the right idé!

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