Higher derivatives of inverse functions (Multivariable Calculus)

Given the function $$ (u,v) = f(x,y) = (x + y, x^2 - y^2) $$ I would like to compute the second partial derivative of $x$ with respect to $v$, at the point $(u,v) = (2,0)$. To calculate the derivative of $f^{-1}$ I know I have to use the fact that the derivative of the $f^{-1}$ at $b$ is the inverse of the $f$ at $a$, where $b = f(a)$ (In this case $(2,0) = f(1,1)$). But I can't go any further than that. Any help is appreciated, thank you.

2 Answers

  1. Jackson- Reply

    2019-11-14

    For the first derivatives (thinking of $x$ and $y$ as functions of $u$ and $v$, and keeping $u$ fixed), you should have $$ \eqalign{0 &= dx + dy\cr dv &= 2 x\; dx - 2 y \; dy\cr}$$ so that $$ \eqalign{\dfrac{\partial x}{\partial v} &= \dfrac{1}{2(x+y)}\cr \dfrac{\partial y}{\partial v} &= \dfrac{-1}{2(x+y)}}$$ Then differentiate the first equation with respect to $y$, using the chain rule. $$ \dfrac{\partial^2 x}{\partial v^2} = - \dfrac{1}{2(x+y)^2} \left(\dfrac{\partial x}{\partial v} + \dfrac{\partial y}{\partial v}\right) = 0$$

  2. Jacob- Reply

    2019-11-14

    In this simple example you can solve the system $$x+y=u,\qquad x^2-y^2=v$$ explicitly for $x$ and $y$. Excluding the degenerate case $u=0$ (which enforces $v=x=y=0$) one finds $$x={1\over2}\left(u+{v\over u}\right),\qquad y={1\over2}\left(u-{v\over u}\right)\ .$$ This immediately leads to $${\partial^2x\over\partial v^2}\equiv0\qquad(u\ne0)\ .$$

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