# Calculus : Chain Rule help

• Feb 18th 2008, 01:32 PM
JonathanEyoon
Calculus : Chain Rule help
I'm asked to find the derivative of the function.

y = R / ( R^2 + 1)^(1/2)

Now I know i'm gonna have to use the quotient rule and incorporate the chain rule in the midst of it. My whole problem is simplifying afterwards. Will someone help me? Thanks

This is what I've got :

((r^2 + 1)^1/2 * 1 - ( r * (1/2)(R^2 + 1)^(-1/2) * 2r)) / r^2 + 1

How do I simplify this so it'll end up looking like

y' = (r^2 + 1)^(-3/2)

• Feb 18th 2008, 01:48 PM
JonathanEyoon
Edit: I didnt figure it out.... any help is appreciated
• Feb 18th 2008, 02:03 PM
Soroban
Hello, Jonathan!

Quote:

Differentiate: . $y \:= \:\frac{r}{(r^2 + 1)^{\frac{1}{2}}}$

My problem is simplifying.

This is what I've got: . $y' \;=\;\frac{(r^2 + 1)^{\frac{1}{2}}\cdot1 - r\cdot\frac{1}{2}(r^2 + 1)^{-\frac{1}{2}}\cdot2r}{r^2 + 1}$ . . . . Correct!

How do I simplify this so it'll end up looking like: . $y' \;= \;(r^2 + 1)^{-\frac{3}{2}}$

$\text{You have: }\;y' \;=\;\frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}}{r^2+1}$

Mutiply top and bottom by $(r^2+1)^{\frac{1}{2}}$

. . $y' \;=\;\frac{(r^2+1)^{\frac{1}{2}}}{(r^2+1)^{\frac{1 }{2}}} \cdot \frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}} {r^2+1} \;=\;\frac{(r^2+1) - r^2}{(r^2+1)^{\frac{3}{2}}} \;=\;\frac{1}{(r^2+1)^{\frac{3}{2}}}$

• Feb 18th 2008, 02:07 PM
JonathanEyoon
Quote:

Originally Posted by Soroban
Hello, Jonathan!

$\text{You have: }\;y' \;=\;\frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}}{r^2+1}$

Mutiply top and bottom by $(r^2+1)^{\frac{1}{2}}$

. . $y' \;=\;\frac{(r^2+1)^{\frac{1}{2}}}{(r^2+1)^{\frac{1 }{2}}} \cdot \frac{(r^2+1)^{\frac{1}{2}} - r^2(r^2+1)^{-\frac{1}{2}}} {r^2+1} \;=\;\frac{(r^2+1) - r^2}{(r^2+1)^{\frac{3}{2}}} \;=\;\frac{1}{(r^2+1)^{\frac{3}{2}}}$

Doh! Question, why multiply by (r^2 +1 ) ^1/2 ?
• Feb 18th 2008, 02:18 PM
angel.white
Quote:

Originally Posted by JonathanEyoon
Doh! Question, why multiply by (r^2 +1 ) ^1/2 ?

Because it will rationalize the numerator:
= $(r^2+1)^{\frac 12+\frac 12}- r^2(r^2+1)^{\frac 12-\frac 12}$

= $(r^2+1)^1- r^2(r^2+1)^0$

= $r^2+1- r^2$

= $+1$

The thing you should notice is that both terms in the numerator have powers of $(r^2+1)^{\frac 12}$ in them. Which upon closer analysis, reveals that you can make the numerator rational by multiplying by $(r^2+1)^{\frac 12}$