Derivative- quotient rule? Need help within hour if possible:)

• Oct 13th 2006, 09:10 PM
thedoge
Derivative- quotient rule? Need help within hour if possible:)
Hi. I'm stuck on a problem I know is fairly simple, but I can not figure out what I am doing wrong.

The problem is f'(x) of f(x)= (sqrt(x)-2)/(sqrt(x)+2)
• Oct 13th 2006, 09:39 PM
earboth
Quote:

Originally Posted by thedoge
Hi. I'm stuck on a problem I know is fairly simple, but I can not figure out what I am doing wrong.

The problem is f'(x) of f(x)= (sqrt(x)-2)/(sqrt(x)+2)

Hi,

I've attached an image to show you what I've calculated.

EB
• Oct 13th 2006, 09:48 PM
thedoge
Thanks mate! Cleared up my issue:)
• Oct 13th 2006, 11:45 PM
CaptainBlack
Quote:

Originally Posted by thedoge
Hi. I'm stuck on a problem I know is fairly simple, but I can not figure out what I am doing wrong.

The problem is f'(x) of f(x)= (sqrt(x)-2)/(sqrt(x)+2)

Now I don't normaly do the quotient rule as it is just a version of the
product rule, but the quotient rule is:

d/dx [g(x)/h(x)] = [h(x)g'(x) - g(x)h'(x)]/[g(x)]^2

In your case g(x)=sqrt(x)-2, and h(x)=sqrt(x)+2, so:

g'(x)=(1/2)/sqrt(x), and h'(x)=(1/2)/sqrt(x).

Therefore applying the quotient rule:

d/dx [sqrt(x)-2)/(sqrt(x)+2)]

.........= [(sqrt(x)+2)/(2 sqrt(x)) - (sqrt(x)-2)/(2 sqrt(x))]/[sqrt(x)+2]^2

Which simplifies to:

d/dx [sqrt(x)-2)/(sqrt(x)+2)] = 2/(sqrt(x)[sqrt(x)+2]^2)

RonL