Implicit Differentiation for a 2nd Derivative

[MATH_IN_A_KHAN]

Hello! As of right now (10:13 PM), I've tried 9 combinations of points to solve this problem. It's a WebWork-based problem that's due in about an hour in a half. Any help would be very, very appreciated.

Homework Statement

I was given this equation: $ln(2y) = 2xy$ and was asked to find the first and second derivative from implicit differentiation.

The first is $2y/(y^-1-2x)$

The second is $4 y^3 (-3+4xy)/(-1+2xy)^3$

These are both correct according to WebWork.

Now, for the final part of the problem, it wants me to find $d^2y/dx^2 = 0$ at $(x,y) = (?, ?)$

The Attempt at a Solution

- 3/(4x) was what I thought the right solution because it makes the numerator equal to zero, but WebWork wants only answers without variables
- Setting $4 y^3 (-3+4xy)/(-1+2xy)^3$ equal to zero and attempting to separate variables. Unsuccessful attempt.
- Messed around with WolframAlpha... also unsuccessful.
- Talked to friends in higher levels of math. No fruits either.

Could someone smarter than I provide some advice?

 

[SammyS]

Hello! As of right now (10:13 PM), I've tried 9 combinations of points to solve this problem. It's a WebWork-based problem that's due in about an hour in a half. Any help would be very, very appreciated.

Homework Statement

I was given this equation: $ln(2y) = 2xy$ and was asked to find the first and second derivative of y from implicit differentiation.

The first derivative is $2y/(y^{-1}-2x)$

The second derivative is $4 y^3 (-3+4xy)/(-1+2xy)^3$

These are both correct according to WebWork.

Now, for the final part of the problem, it wants me to find $d^2y/dx^2 = 0$ at $(x,y) = (?, ?)$

The Attempt at a Solution

- 3/(4x) was what I thought the right solution because it makes the numerator equal to zero, but WebWork wants only answers without variables
- Setting $4 y^3 (-3+4xy)/(-1+2xy)^3$ equal to zero and attempting to separate variables. Unsuccessful attempt.
- Messed around with WolframAlpha... also unsuccessful.
- Talked to friends in higher levels of math. No fruits either.

Could someone smarter than I provide some advice?

I'll give it a whirl anyway.


You're correct in that y'' = 0 when y = -3/(4x) .

Parametrize that. Let x = a, then y'' = 0, at (x,y) = (a, -3/(4a)) .


(There is another set of solutions too.)

 

[MATH_IN_A_KHAN]

I'll give it a whirl anyway.


You're correct in that y'' = 0 when y = -3/(4x) .

Parametrize that. Let x = a, then y'' = 0, at (x,y) = (a, -3/(4a)) .


(There is another set of solutions too.)

Hmmm... no dice.

I tried (a, -3/(4a)), (1, -3/(4)), (2, -3/(8)), (-1, -3/(-4)), etc. Didn't work.

Here's a screen cap of the problem - http://i.imgur.com/k5XsDH6.png - if that can be of any help.

 

[SammyS]

Hmmm... no dice.

I tried (a, -3/(4a)), (1, -3/(4)), (2, -3/(8)), (-1, -3/(-4)), etc. Didn't work.

Here's a screen cap of the problem - http://i.imgur.com/k5XsDH6.png - if that can be of any help.

Of course, with (a, -3/(4a)), you must not include x = 0 , i.e. you can't have a = 0.

Also, what is the second derivative if y = 0 , no matter the value of x ?