Second Derivative

[math812]

for the equation x^1/2 + y^1/2 = 1, i was able to find the first derivative as follows:

(-y^1/2) / (x^1/2)

however, from here i can not seem to get the second derivative correct using the quotient rule, can anyone offer suggestions?

thanks

 

[Unco]

Are you familiar with implicit differentiation?

d/dx -y^(1/2) = -(1/2)y^(-1/2)(dy/dx)

... and that kind of stuff?

 

[math812]

yes, i know about implicit differentiation; so far i have solved this much:

-1/2rooty * dy/dx * rootx - (-rooty * 1/2rootx)
------------------------------------------------------
(rootx)^2

since dy/dx = -rooty/rootx, it can cancel that part of the problem, leaving:

1/2 + rooty/2rootx
----------------------
x

and then i mult. by 2root x, giving:

rootx + rooty
----------------
2xrootx

however, the answer in the back says that the denominator should be 2x^3/2

where am i going wrong?[/code]

 

[math812]

nevermind, i see how the exponents in x^1 * x^1/2 add to 3/2

sorry!

 

[soroban]

Hello, math812!

Hey, you did great!

They usually simplify the answer even further . . .

. . . . . . . . . . . . . . . .x^1/2 + y^1/2
You have: . y" . = . -------------- . which is absolutely correct.
. . . . . . . . . . . . . . . . . 2x^3/2

But the original equation told us that: .x^1/2 + y^1/2 .= .1

. . . . . . . . . . . . . . . . . . . . . . .1
So the answer is: . y" . = . -------
. . . . . . . . . . . . . . . . . . . . . 2x^3/2

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

This final substitution turns up often with these problems
. . (second derivatives of implicit functions).

It is something to look for. .Many teachers and authors
. .are not even aware that this is possible.