1. ## Implicit function

Hey.
I have this problem with the second part of the question.
In the first part, I needed to find all the extrema points of this implicit function and the second one is to find if it's minimum or maximum.
I posted only the derivative of the implicit function and the points I found.
BTW there was another condition to find the point, they said F(a) = b, so together with the derivative, I found the points.
Back to the question, as I was saying, the second part is to find if it's minimum or maximum, how do I do that?
I thought about finding the second derivative and plugging the points into it, but how do you find the second derivative of an implicit function, same as you find the first one?

I didn't posted the question, cause it's kind hard to translate, so I hope I gave you all the info you need.

2. Originally Posted by asi123
I thought about finding the second derivative and plugging the points into it, but how do you find the second derivative of an implicit function, same as you find the first one?
Yes, but it can get a little more complex...and messy... >_>

$\frac{\,dy}{\,dx}=\frac{1-x+2y}{16y^2-8xy+2}$

Therefore $\frac{d^2y}{dx^2}=\frac{\left(16y^2-8xy+2\right)\left(-1+2\frac{\,dy}{\,dx}\right)-\left(1-x+2y\right)\left(32y\frac{\,dy}{\,dx}-8y-8x\frac{\,dy}{\,dx}\right)}{(16y^2-8xy+2)^2}$

..and then wherever you see $\frac{\,dy}{\,dx}$, plug in $\frac{1-x+2y}{16y^2-8xy+2}$...

Its gonna get messy...

I hope this helps.

--Chris

3. Originally Posted by Chris L T521
Yes, but it can get a little more complex...and messy... >_>

$\frac{\,dy}{\,dx}=\frac{1-x+2y}{16y^2-8xy+2}$

Therefore $\frac{d^2y}{dx^2}=\frac{\left(16y^2-8xy+2\right)\left(-1+2\frac{\,dy}{\,dx}\right)-\left(1-x+2y\right)\left(32y\frac{\,dy}{\,dx}-8y-8x\frac{\,dy}{\,dx}\right)}{(16y^2-8xy+2)^2}$

..and then wherever you see $\frac{\,dy}{\,dx}$, plug in $\frac{1-x+2y}{16y^2-8xy+2}$...

Its gonna get messy...

I hope this helps.

--Chris

Why do I need to plug in the all expression?
I mean, I have the extrema points.

4. Originally Posted by asi123
Why do I need to plug in the all expression?
I mean, I have the extrema points.
Ah... good point.

Plug in the points of the extrema into the messy second derivative here:

$\frac{d^2y}{dx^2}=\frac{\left(16y^2-8xy+2\right)\left(-1+2\frac{\,dy}{\,dx}\right)-\left(1-x+2y\right)\left(32y\frac{\,dy}{\,dx}-8y-8x\frac{\,dy}{\,dx}\right)}{(16y^2-8xy+2)^2}$

where $\frac{\,dy}{\,dx}=\frac{1-x+2y}{16y^2-8xy+2}$

At $(1,0)$, we get

$\frac{d^2y}{dx^2}=\frac{\left(2\right)\left(-1\right)}{(2)^2}=\color{red}\boxed{-\frac{1}{2}}<0$

This implies that there is a maximum at $(1,0)$

I'll leave the other one for you.

--Chris