Find derivative using Inverse Function Theorem

Question

If $f(u,v)=(3u^2-v,2u^4+v^3)$ then show that $f^{-1}$ exists and is differentiable in some nbd. of $(1,2)$ and find out $D(f^{-1})(1,2)$.

Clearly $f$ is contnuously differentiable and we get, $$Df(1,2)=\left(\begin{matrix}6 & -2\\8&12\end{matrix}\right)\ne \mathcal O.$$ So by inverse function theorem, $f^{-1}$ exists and differentiable in nbd, of $(1,2)$.

Again we have, $Df^{-1}(b)=\frac{1}{Df(a)}$ , where $f(a)=b.

Now $f(u,v)=(1,2)$ gives $\displaystyle \begin{cases}u=-0.82\\v=1.03\end{cases}$ and $\displaystyle \begin{cases}u=0.82\\v=1.03\end{cases}$.

Therefore $Df^{-1}(1,2)=\frac 1{Df(-0.82,1.03)}$ and $Df^{-1}(1,2)=\frac 1{Df(0.82,1.03)}$. We get two different values!! Can we say derivative exists? Where my mistake ?

Answer 1

You're skipping over what the inverse function theorem says. It doesn't say $f^{-1}$ exists in a neighborhood of $(1,2)$ (well that's part of it, but not quite right). Also, you haven't correctly verified that the hypotheses of the IFT are satisfied. What it says is the following:

if $f$ is $C^r$ and $(u_0,v_0)$ is a point in the domain of $f$ such that $f(u_0,v_0)=(1,2)$ and $Df_{(u_0,v_0)}$ is invertible (note that this is a much stronger requirement in dimensions $\geq 2$ than simply requiring the derivative to be non-zero), then there exist open neighborhoods $A$ of $(u_0,v_0)$ and $B$ of $f(u_0,v_0)=(1,2)$ such that $f$ maps $A$ bijectively onto $B$ and the (locally defined) inverse $\phi:B\to A$ is $C^r$ (with such and such formula for the derivative).

Note that I'm hesitant to use the $f^{-1}$ notation because you seem to be confusing what it means to have a local inverse. Here, $f$ is a mapping $\Bbb{R}^2\to\Bbb{R}^2$. The inverse first of all need not even exist, and even if it doesn, it's only defined on the image of $f$.

So your mistakes are the following:

• You simply showed $Df(1,2)\neq 0$. This is NOT at all what was required by the theorem. Furthermore, it seems like you're confusing the role of the domain and target space. Based on how you phrase the question, we should be thinking of $(1,2)$ as being an element of the target space so it shouldn't even try to evaluate the derivative $Df(1,2)$.

• As you can see from the statement of the theorem, it is a purely local assertion, and even more importantly, YOU have to first specify the point $(u_0,v_0)$ which you're interested in. Also, I doubt your calculations are correct when solving the equation $f(u,v)=(1,2)$; a simple check in the calculator shows that $3(0.82)^2-1.03=0.9872\neq 1$, so maybe you used some approximation? In that case it is very misleading to put an equal sign in your second last line.

• Also, derivatives in dimension $n\geq 2$ are linear transformations $\Bbb{R}^n\to\Bbb{R}^n$ (or equivalently, once you choose a basis... usually the "standard" basis), you can think of this as an $n\times n$ matrix. So, using the fraction notation $\frac{1}{Df(\text{evaluated at some point})}$ is terrible notation. Instead use the matrix inversion notation $[Df(\text{evaluated at some point})]^{-1}$.

Of course, purely by looking at the equations, I can tell that if $(u_0,v_0)$ satisfies $f(u_0,v_0)=(1,2)$ then $f(-u_0,v_0)=(1,2)$ as well. Now, what you have to do is fix a particular $(u_0,v_0)$ such that $f(u_0,v_0)=(1,2)$. After this, show that $Df_{(u_0,v_0)}$ is invertible. Then, apply the IFT to get open neighborhoods $A$ of $(u_0,v_0)$ and $B$ of $(1,2)$ along with a $C^{\infty}$ local inverse $\phi:B\to A$ of $f:A\to B$. Then, calculate $D\phi_{(1,2)}$.

This is not contradictory at all, because if $(-u_0,v_0)$ is the other point which satisfies $f(-u_0,v_0)=(1,2)$, then after verifying that $Df_{(-u_0,v_0)}$ is invertible, you can apply the IFT to get (in general different than above) open neighborhoods $A'$ of $(-u_0,v_0)$ and $B'$ of $(1,2)$ along with a $C^{\infty}$ local inverse $\psi:B'\to A'$ of $f:A'\to B'$. Now, you can calculate $D\psi_{(1,2)}$.

Hopefully, from my emphasis on the different notation, it's clear that you're dealing with a local situation, with two different functions $\phi,\psi$, both of which you conveniently (but albeit confusingly if this is your first time learning this stuff) simply by $f^{-1}$. I guess if you don't want to introduce new letters, then the notation $(f|_A)^{-1}$ and $(f|_{A'})^{-1}$ also works.

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For example, this situation is analogous to the simple trig functions from single variable calculus. If $f(x)=\sin x$, then the restrictions

• $f|_{[-\frac{\pi}{2},\frac{\pi}{2}]}:[-\frac{\pi}{2},\frac{\pi}{2}]\to [-1,1]$ and
• $f|_{[\frac{\pi}{2},\frac{3\pi}{2}]}:[\frac{\pi}{2},\frac{3\pi}{2}]\to [-1,1]$

are both invertible, but of course their inverses are completely different functions.