# Derivate

• Feb 20th 2013, 04:56 AM
Petrus
Derivate
Hello!I got big problem derivate $\displaystyle \frac{1}{sqrt(x)}$
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 $\displaystyle \frac{f(a+h)-f(a)}{h}$
• Feb 20th 2013, 05:10 AM
Prove It
Re: Derivate
\displaystyle \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 \displaystyle \begin{align*} &= \frac{-1}{\sqrt{x} \sqrt{x} \left( \sqrt{x} + \sqrt{x} \right) } \\ &= \frac{-1}{2x\sqrt{x}} \end{align*}
• Feb 20th 2013, 05:11 AM
MathJack
Re: Derivate
1/(sqrt(x) + h) - 1/(sqrt(x))
multiply top and bottom by conjugate
then let h = 0
• Feb 20th 2013, 05:17 AM
Petrus
Re: Derivate
I see what i did wrong on My progress!! Ty alot:) Well i had the right idé!