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So I think I understand how to use the Implicit Function theorem to find partial derivatives given one function but I am confused as to how to do this for 2 functions. I'm trying to find $\frac{\partial x}{\partial u}, \frac{\partial x}{\partial v}$ around the point $(1,-1,-1,2)$ given the equations $$x^2+2y^2+u^2+v=6$$$$2x^3+4y^2+u+v^2=9$$ and I thus calculated that $$Df(1,-1,-1,2)=\begin{bmatrix} 2 & -4 & -2 & 1\\6 & -8 & 1 & 4\end{bmatrix}$$ but I'm not sure where to go from here since the equation for the partial of the implicit function in my textbook is only for cases where we have one equation? Thanks in advance for any help!

EDIT: So I've looked at the problem a bit further and I'm still confused as to how the derivative for an implicit function works in the case of $2+$ equations? Could anyone give an example with simpler equations so I could then apply it to this system? Thanks again.

Twis7ed
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    Can you please state the IFT as you know it? – Git Gud Mar 04 '16 at 23:09
  • We just learned it as: If $f\colon \mathbb{R}^m\times\mathbb{R}^n\to \mathbb{R}^m$ s.t. $f\in\mathscr{C}^1$ and we fix $(x_0,y_0)$ such that $f(x_0,y_0)=0$ and our derivative matrix is nonsingular then there exist open sets $U\subset\mathbb{R}^n, V\subset\mathbb{R}^m$ such that in those sets we can define a $\phi(x)\colon U\to V$ which is $\mathscr{C}^1$ but for this multivariate version we don't get what the derivatives should be, we only get such an equation for the $\mathbb{R}^n\to\mathbb{R}$ case. – Twis7ed Mar 05 '16 at 15:08
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    This formulation of the implicit function theorem directly gives you the derivatives. And in this answer you can see an example of a derivative found for a one equation implicit differentiation. – Git Gud Mar 06 '16 at 12:19
  • Thank you so much! That link is very helpful. – Twis7ed Mar 06 '16 at 12:26
  • You're welcome. I realise this is a lot to take in. If you don't understand something, just ask. – Git Gud Mar 06 '16 at 22:54

1 Answers1

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You want to find $\dfrac{\partial x}{\partial u}(-1,2)$ and $\dfrac{\partial x}{\partial v}(-1,2)$. This really doesn't make sense because $x$ isn't a function, it is a variable. I, however, know what is meant and I will translate the problem into something I can understand.

In the notation of the formulation of the IFT you provided, set $m=2=n, x_0=(1,-1)$ and $y_0=(-1,2)$. Your $\phi(x)$, however, will be a "$\phi(y)$", that is, the first two variables will be determined by the last two.

Let $$f_1\colon \mathbb R^4\to\mathbb R, (x,y,u,v)\to x^2+2y^2+u^2+(-6),$$ $$f_2\colon \mathbb R^4\to\mathbb R, (x,y,u,v)\to 2x^3+4y^2+u+v^2-9$$ and $F=(f_1, f_2)$.

Since $F(1,-1,-1,2)=(0,0)$ and $$\begin{bmatrix} (\partial_xf_1) & (\partial_yf_1)\\ (\partial_xf_2) & (\partial_yf_2)\end{bmatrix}_{(1,-1,-1,2)}=\ldots= \begin{bmatrix} 2 & -4\\6 & -8\end{bmatrix}$$ with $\begin{bmatrix} 2 & -4\\6 & -8\end{bmatrix}$ being invertible, you can apply the IFT. (Here $\partial_xf_1$ denotes the derivative of $f_1$ with respect to the first variable. It's similar to the remaining partials derivatives).

Using the version of the IFT you're familiar with, you get the existence of a function $G\colon U\to V$ in $\mathscr{C}^1$ (where $U$ and $V$ are neighborhoods of $(-1,2)$ and $(1,-1)$ respectively) such that $G(-1,2)=(1,-1)$ and $\color{blue}{F(g_1(u,v), g_2(u,v),u,v))=(0,0)}$, for all $(u,v)\in U$, where $g_1,g_2$ are functions such that $G=(g_1, g_2)$. (In the notation of your IFT, $G=\phi$). (You're specially interested in $g_1$, it is what I guess you'd call $x(u,v)$).

Let $\psi\colon U\to\mathbb R^4, (u,v)\to (g_1(u,v), g_2(u,v),u,v)$ and $H=F\circ\psi$. This denomination allows one to rewrite the blue condition above as $H(u,v)=(0,0)$, for every $(u,v)\in U$.

Since $H$ is null, its Jacobian matrix $DH$ will be the null linear map and using the chain rule on $H\color{grey}{=F\circ \psi}$ yields $$\begin{align} \color{grey}{\begin{bmatrix}0&0\\0 & 0 \end{bmatrix}=}(DH)(u,v)&=(DF)(\psi(u,v))(D\psi)(u,v)\\ &=\begin{bmatrix} 2g_1 & 4g_2 & 2u & 1\\ 6g_1^2 & 8g_2 & 1& 2v\end{bmatrix}_{(u,v)}\begin{bmatrix}(\partial _ug_1) & (\partial_vg_1)\\ (\partial_u g_2) & (\partial_v g_2)\\ 1& 0\\ 0 & 1 \end{bmatrix}_{(u,v)}\\ &=\begin{bmatrix} 2g_1(\partial_u g_1) + 4g_2(\partial _ug_2)+2u & 2g_1\partial_vg_1 + 4g_2(\partial_v g_2)+1\\6g_1^2(\partial_u g_1)+8g_2(\partial _ug_2)+1 & 6g_1^2\partial_vg_1+8g_2(\partial_v g_2)+2v\end{bmatrix}_{(u,v)} \end{align}$$

This gives you a system of equations which you should be able to solve.

For instance to find $(\partial _ug_1)(-1,2)$ you can use the entries $(1,1)$ and $(2,1)$ of the matrix equality above to get $$\begin{cases}2g_1(-1,2)(\partial_u g_1)(-1,2)+4g_2(-1,2)(\partial_u g_2)(-1,2)-2&=0\\ 6(g_1(-1,2))^2(\partial_u g_1)(-1,2)+8g_2(-1,2)(\partial_u g_2)(-1,2)+1&=0 \end{cases}$$ and consequently (due to $g_1(-1,2)=1$ and $g_2(-1,2)=-1$), $$\begin{cases}2(\partial_u g_1)(-1,2)-4(\partial_u g_2)(-1,2)-2&=0\\ 6(\partial_u g_1)(-1,2)-8(\partial_u g_2)(-1,2)+1&=0 \end{cases}_.$$

Multiplying the first row by $-2$ and adding to the second one ultimately yields $(\partial_u g_1)(-1,2)=-\dfrac 5 2$.

You can find $(\partial_v g_1)(-1,2)$ yourself. You can confirm the result with what you get from direct application of the version of the IFT stated here.

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