Thread: 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)


Thanks in advance

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

Hello, Jonathan!

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

My problem is simplifying.

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

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

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

. . $\displaystyle 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}}} $

Originally Posted by Soroban

Hello, Jonathan!

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

. . $\displaystyle 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 ?

Originally Posted by JonathanEyoon

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

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

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

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

= $\displaystyle +1$

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