# Second derivatives using implicit differentiation

1. Oct 17, 2010

### 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 gonna 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 wanna find $$\frac{{\partial^2 x}}{{\partial y^2}}$$ how should I proceed?

2. Oct 17, 2010

### Staff: Mentor

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.

3. Oct 18, 2010

### 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".

4. Oct 18, 2010

Thanks!