Second derivatives using implicit differentiation

[Telemachus]

Hi there. Well, I wanted to know how to find the second derivatives of a function using implicit differentiation. Is it possible? I think it is. I think I must use the chain rule somehow, but I don't know how... I'm in multivariable calculus since the function I'm going to use could be seen as a function of only one variable.

An ellipse: F(x,y)=4x^2+y^2-25=0

So we have the partial derivatives:
F_x=8x, F_y=2y
F_{xx}=8, F_{yy}=2

So then, using implicit differentiation:

\frac{{\partial x}}{{\partial y}}=-\displaystyle\frac{\frac{{\partial F}}{{\partial y}}}{\frac{{\partial F}}{{\partial x}}}=\displaystyle\frac{-y}{4x}

But now if I want to find \frac{{\partial^2 x}}{{\partial y^2}} how should I proceed?

 

[Mark44]

I believe you're going about this in the wrong way, since partial derivatives are not required.

If 4x² + y² + 25 = 0,
then 8x + 2y*y' = 0 ==> y' = -4x/y

To get y'', simply take the derivative with respect to x of -4x/y, using the quotient rule and the chain rule.

 

[HallsofIvy]

Or, much the same way, but, I think, a bit more in keeping with the spirit of "implicit differentiation", from 4x^2+ y^2+ 25= 0, 8x+ 2y y'= 0 and differentiating again, 8+ 2y'(y')+ 2yy"=8+ 2y'^2+ 2yy"= 0 and solve that for y".

 

[Telemachus]

Thanks!