First principles

• October 30th 2009, 01:04 PM
sammy28
First principles
hi all,

Find $\frac{dy}{dx}$ for $y=\sqrt{x}$

$= \frac{1}{2}\displaystyle{x}^{-\frac{1}{2}}\equiv\frac{1}{2\sqrt{x}}$

i then try to get this result from first principles:

$f'(x)=\lim_{h\rightarrow0}\frac{f(x+h)-f(x)}{h}=\lim_{h\rightarrow0}\frac{(x+h)^{\frac{1} {2}} - x^{\frac{1}{2}}}{h}$

then cancel the fractional powers by raising both numerator and denominator by reciprocal power, so that

$\lim_{h\rightarrow0}\frac{h}{h^2}=0$

what am i doing wrong?

thanks sammy.
• October 30th 2009, 01:38 PM
TwistedOne151
$(\sqrt{x+h}-\sqrt{x})^2\ne(x+h)-x$. To find the limit, we rationalize the numerator
$\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+ \sqrt{x}}$
$=\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$
$=\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$
$=\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x})}$,

--Kevin C.
• October 30th 2009, 02:37 PM
sammy28
thanks a lot for that (Clapping).
i should have tried expanding the numerator and would have seen it was wrong! (Giggle)