Question

f(x)= 1/Sqrt(x) , Find using definition of Derivative
F(x)=?

Answer 1

not really i dont really understand it...its was a homework problem assigned

Answer 2

the "definition" is:\[\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}\]

Answer 3

\[f(x)=\frac{1}{\sqrt x}\Rightarrow f'(x)=\lim_{h\to0}\frac{\frac{1}{\sqrt{x+h}}-\frac{1}{\sqrt x}}{h}\]

Answer 4

From there it is all just simplification

Answer 5

is this the answer

Answer 6

ow my eyes!!!

Answer 7

LOL! My eyes they burn!!!!!!!

Answer 8

\[\frac{\frac{1}{.x+h}-\frac{1}{.x}}{h}\]
\[\frac{\frac{.x-.x+h}{.x.x+h}}{h}\]
\[\frac{.x-.x+h}{h({.x.x+h})}\]
\[\frac{.x-.x+h}{h({.x.x+h})}*\frac{.x+.x+h}{.x+.x+h}\]
\[\frac{x-(x+h)}{h({.x.x+h}){(.x+.x+h)})}\]
\[\frac{h}{h({.x.x+h}){(.x+.x+h)})}\]
\[\frac{1}{({.x.x+h}){(.x+.x+h)})}\]
and when h=0
\[\frac{1}{({.x.x}){(.x+.x})}\]
\[\frac{1}{x{2\sqrt{x}}}\]
\[\frac{1}{{2x~\sqrt{x}}}\]

Answer 9

so yes, that looks fine :)

Answer 10

or simply:\[\frac12x^{-3}\]