# Second derivatives using implicit differentiation

• Oct 17th 2010, 02:47 PM
Ulysses
Second derivatives using implicit differentiation
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: $\displaystyle F(x,y)=4x^2+y^2-25=0$

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

So then, using implicit differentiation:

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

But now if I wanna find $\displaystyle \frac{{\partial^2 x}}{{\partial y^2}}$ how should I proceed?
• Oct 17th 2010, 02:56 PM
lvleph
Well, if you actually want to find $\displaystyle \frac{\partial^2 x}{\partial y^2}$ and not $\displaystyle \frac{\partial^2 y}{\partial x^2}$, then you will just differentiate the derivative you already have, i.e.
$\displaystyle \frac{\partial^2 x}{\partial y^2} = \frac{\partial x}{\partial y}\left[-\frac{y}{4x}\right]$
• Oct 17th 2010, 03:05 PM
Ulysses
The thing is that it gives $\displaystyle -\frac{1}{4x}$, and If we differentiate expressing x=x(y)

Look, this is the second derivative using the explicit form: Clic here
• Oct 17th 2010, 03:17 PM
lvleph
I do this differently than you, so hopefully I won't confuse you.
Here I am assuming that x is a function of y.
$\displaystyle \frac{d}{dy}\left[4x^2+y^2-25\right]=\frac{d}{dy}\left[0\right]$
$\displaystyle 8xx'+2y=0 \Rightarrow x' = -\frac{y}{4x}$
$\displaystyle \frac{d}{dy}\left[x'\right] = -\frac{d}{dy}\left[\frac{1}{4}yx^{-1}\right]$
$\displaystyle x'' = -\frac{1}{4}x^{-1} + \frac{1}{4}yx^{-2} x'$
Now we substitute the formula for $\displaystyle x'$
$\displaystyle x'' = -\frac{1}{4}x^{-1} + \frac{1}{4}yx^{-2} \left(-\frac{y}{4x}\right)$
$\displaystyle x'' = -\frac{1}{4}x^{-1} - \frac{y^2}{16x^3} \Rightarrow x'' = \frac{-4x^2 - y^2}{16x^3}$.
• Oct 17th 2010, 03:30 PM
Ulysses
Great explanation.