# derivatives - double check my work please

• May 22nd 2006, 02:30 PM
becky
derivatives - double check my work please
Please double check my work - just doesn't 'feel' right and I've always had a problem with what I call 'three stacked fractions'.
Instructions: Compute the derivative of the given function .

h(u) = 1/square root of u

My solution: h'(u) lim h->0 1/square root x+h - 1/square root u/h=

square root u/x+h - square root x+h/u /h(square root x+h)(square root u)=

square root u(x+h) - square root x+h(x+h/h(square root u)(x+h)^2=

-square root x.
• May 22nd 2006, 02:52 PM
ThePerfectHacker
Quote:

Originally Posted by becky
Please double check my work - just doesn't 'feel' right and I've always had a problem with what I call 'three stacked fractions'.
Instructions: Compute the derivative of the given function .

h(u) = 1/square root of u

My solution: h'(u) lim h->0 1/square root x+h - 1/square root u/h=

square root u/x+h - square root x+h/u /h(square root x+h)(square root u)=

square root u(x+h) - square root x+h(x+h/h(square root u)(x+h)^2=

-square root x.

You should get for, $\displaystyle \frac{1}{\sqrt{u}}$ the answer of,
$\displaystyle 2\sqrt{u}$
• May 22nd 2006, 11:19 PM
CaptainBlack
Quote:

Originally Posted by ThePerfectHacker
You should get for, $\displaystyle \frac{1}{\sqrt{u}}$ the answer of,
$\displaystyle 2\sqrt{u}$

$\displaystyle \frac{d}{du} \left[\frac{1}{\sqrt{u}}\right]=-\frac{1}{2}\frac{1}{(\sqrt{u})^3}$

RonL
• May 23rd 2006, 01:19 PM
ThePerfectHacker
Quote:

Originally Posted by CaptainBlack
$\displaystyle \frac{d}{du} \left[\frac{1}{\sqrt{u}}\right]=-\frac{1}{2}\frac{1}{(\sqrt{u})^3}$

RonL

Sorry :o
I took the integral