# Derivatives of a square

• Oct 2nd 2006, 06:52 PM
cyberdx16
Derivatives of a square
i have here (sq root of s)-1/(sq root of s)+1 how do i find the derivative of this??? i've been getting all these weird 1/2s and what not and cant figure out what im doing wrong
• Oct 2nd 2006, 07:40 PM
CaptainBlack
Quote:

Originally Posted by cyberdx16
i have here (sq root of s)-1/(sq root of s)+1 how do i find the derivative of this??? i've been getting all these weird 1/2s and what not and cant figure out what im doing wrong

Lets assume that your function is:

h(s)=(sqrt(s)-1)/(sqrt(s)+1).

Now you probably need to use the quotient rule here, but I don't
know the quotient rule so I will use the product rule:

d/ds[f(s)/g(s)]=[df/ds]/g(s)-f(s)[dg/ds]/(g(s))^2.

In this case f(s)=sqrt(s)-1, so df/ds=1/(2 sqrt(s)), and g(s)=sqrt(s)+1,
so dg/ds=1/(2 sqrt(s), so we have:

dh/ds=1/[2 sqrt(s) (sqrt(s)+1)] - (sqrt(s)-1)/[2 sqrt(s) (sqrt(s)+1)^2]

which simplifies down to:

dh/ds=1/[sqrt(s) (sqrt(s)+1)^2]

(which is much easier to see if you write it out on paper (or if the
mathematical type setting system we used to have here was working)

RonL