derivatives - double check my work please

**becky** · May 22nd 2006, 02:30 PM

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.

**ThePerfectHacker** · May 22nd 2006, 02:52 PM

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, $\frac{1}{\sqrt{u}}$ the answer of,
$2\sqrt{u}$

**CaptainBlack** · May 22nd 2006, 11:19 PM

Originally Posted by ThePerfectHacker

You should get for, $\frac{1}{\sqrt{u}}$ the answer of,
$2\sqrt{u}$

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

RonL

**ThePerfectHacker** · May 23rd 2006, 01:19 PM

Originally Posted by CaptainBlack

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

RonL

Sorry

I took the integral

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