Calculus : Chain Rule help

**JonathanEyoon** · February 18th 2008, 01:32 PM

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)

Thanks in advance

**JonathanEyoon** · February 18th 2008, 01:48 PM

Edit: I didnt figure it out.... any help is appreciated

**Soroban** · February 18th 2008, 02:03 PM

Hello, Jonathan!

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 $(r^2+1)^{\frac{1}{2}}$

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

**JonathanEyoon** · February 18th 2008, 02:07 PM

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 ?

**angel.white** · February 18th 2008, 02:18 PM

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 $(r^2+1)^{\frac 12}$

Math Help - Calculus : Chain Rule help

LinkBack

Thread Tools

Display

Calculus : Chain Rule help

Similar Math Help Forum Discussions

Calculus chain rule questions

Multivariable Calculus - Chain Rule

Calculus chain rule

calculus: the chain rule

Calculus - Chain Rule Derivative

Search Tags