Originally Posted by
lilheadbaby1 Find the derivate of p(x)= (3x)^1/2
using the alternative formula for the derivate
f ' (x)= lim f(z)-f(x)/z-x
z>x
my answer was 1/2(3x)^1/2
but its wrong
$\displaystyle \frac{f(z)-f(x)}{z-x}=\frac{(3z)^{1/2}-(3x)^{1/2}}{z-x}$
$\displaystyle
=(3z)^{1/2}\frac{(1-(x/z)^{1/2})}{z(1-x/z)}
$
$\displaystyle
=\left( \frac{3}{z} \right)^{1/2} \frac{1-(x/z)^2}{(1+(x/z)^{1/2})(1-(x/z)^{1/2})}$
$\displaystyle
=\left( \frac{3}{z} \right)^{1/2} \frac{1}{(1+(x/z)^{1/2})}$
So:
$\displaystyle
\lim_{z \rightarrow x}\frac{f(z)-f(x)}{z-x}=\left( \frac{3}{x} \right)^{1/2} \frac{1}{2}= \frac{\sqrt{3}}{2} x^{-1/2}
$
RonL