1. ## First principles

hi all,

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

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

i then try to get this result from first principles:

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

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

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

what am i doing wrong?

thanks sammy.

2. $\displaystyle (\sqrt{x+h}-\sqrt{x})^2\ne(x+h)-x$. To find the limit, we rationalize the numerator
$\displaystyle \lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim_{h\to0}\frac{\sqrt{x+h}-\sqrt{x}}{h}\frac{\sqrt{x+h}+\sqrt{x}}{\sqrt{x+h}+ \sqrt{x}}$
$\displaystyle =\lim_{h\to0}\frac{(x+h)-x}{h(\sqrt{x+h}+\sqrt{x})}$
$\displaystyle =\lim_{h\to0}\frac{h}{h(\sqrt{x+h}+\sqrt{x})}$
$\displaystyle =\lim_{h\to0}\frac{1}{(\sqrt{x+h}+\sqrt{x})}$,

--Kevin C.

3. thanks a lot for that .
i should have tried expanding the numerator and would have seen it was wrong!