Let f:Rn - Rn and g:Rn - Rn be inverse functions. Suppose f and g are differentiable. Show that determianants of Df and Dg are not zero, and that Dg=(Df)^-1. Where Df and Dg is the derivative of f and g respectively.

Question

Let f:Rn -> Rn and g:Rn -> Rn be inverse functions. Suppose f and g are differentiable. Show that determianants of Df and Dg are not zero, and that Dg=(Df)^-1. Where Df and Dg is the derivative of f and g respectively.

cruffo · Answer

The "Df and Dg are not zero" part should come directly from the definition of differentiable.  However, I don't see how "Dg=(Df)^-1" could be possible in general.  Are you given any more info on f and g?

cruffo · Answer

Ahhh..  f and g are inverses of each other...

anonymous · Answer

(Df)^-1 is the inverse matrix of Df btw

cruffo · Answer

yep.  I know the notation.

anonymous · Answer

that is all the info given and I'm not sure how the not zero part follows

anonymous · Answer

i mean i understand that Df and D(f^-1)=Dg exist but unless I can show D(f^-1) = (Df)^-1 then i cannot say that since they are invertible the determinant is nonzero...

cruffo · Answer

First paragraph of the following http://people.math.gatech.edu/~ghomi/LectureNotes/LectureNotes6G.pdf may provide the idea of how to prove that the Jacobian is nonsingular. I'll look back at my DG book.

anonymous · Answer

thx for the tip, gona take a look now....

cruffo · Answer

I got the following
Since $f \circ g = g \circ f = I(R_n)$, 
then $D(f \circ g) = Df_{|g} \cdot Dg = D I(R_n)$.  
So $|Df_{|g} \cdot Dg | = |Df_{|g}| \cdot |Dg | = |D I(R_n)|$.  
But $ |D I(R_n)| = 1$.  
Thus  $|Dg | \neq 0$.

Similiar proof for $|Df|$.