Part (a): Proving Local Bijectivity of ( g ) near ( (0,1) ) #

Let's start by leaning on the Inverse Function Theorem—this is exactly the tool we need here. The theorem tells us that if a function is differentiable at a point and its Jacobian matrix at that point is non-singular (determinant ≠ 0), then the function maps a neighborhood of that point bijectively to a neighborhood of its image.

First, we already have the Jacobian matrix for ( g ):
[
Dg(x,y) = \begin{bmatrix} 4ye^{2x} & 2e^{2x} \ e^y & xe^y \end{bmatrix}
]

Plug in ( (x,y) = (0,1) ):
[
Dg(0,1) = \begin{bmatrix} 4(1)e^{0} & 2e^{0} \ e^1 & (0)e^1 \end{bmatrix} = \begin{bmatrix} 4 & 2 \ e & 0 \end{bmatrix}
]

Now calculate its determinant:
[
\det(Dg(0,1)) = (4)(0) - (2)(e) = -2e
]
Since ( -2e \neq 0 ), the Jacobian is non-singular. Also, ( g ) is clearly differentiable everywhere (all its component functions are smooth, infinitely differentiable, over ( \mathbb{R}^2 )).

By the Inverse Function Theorem, ( g ) maps some neighborhood of ( (0,1) ) one-to-one and onto some neighborhood of ( g(0,1) = (2(1)e^{0}, 0 \cdot e^1) = (2,0) ). That's exactly what we needed to prove.

Part (b): Computing ( D(f \circ g^{-1}) ) at ( (2,0) ) #

We'll use the Chain Rule combined with the Inverse Function Theorem here. Remember that:
[
D(f \circ g^{-1})(z) = Df(g^{-1}(z)) \cdot Dg^{-1}(z)
]
And from the Inverse Function Theorem, ( Dg^{-1}(z) = \left[ Dg(g^{-1}(z)) \right]^{-1} ).

First, note that ( g(0,1) = (2,0) ), so ( g^{-1}(2,0) = (0,1) )—this is our key reference point.

Step 1: Compute ( Df(0,1) ) #

The function ( f(x,y) = (3x - y^2, 2x + y, xy + y^3) ), so its Jacobian matrix is:
[
Df(x,y) = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \ \frac{\partial f_3}{\partial x} & \frac{\partial f_3}{\partial y} \end{bmatrix} = \begin{bmatrix} 3 & -2y \ 2 & 1 \ y & x + 3y^2 \end{bmatrix}
]

Substitute ( (x,y) = (0,1) ):
[
Df(0,1) = \begin{bmatrix} 3 & -2(1) \ 2 & 1 \ 1 & 0 + 3(1)^2 \end{bmatrix} = \begin{bmatrix} 3 & -2 \ 2 & 1 \ 1 & 3 \end{bmatrix}
]

Step 2: Compute ( \left[ Dg(0,1) \right]^{-1} ) #

We already have ( Dg(0,1) = \begin{bmatrix} 4 & 2 \ e & 0 \end{bmatrix} ). For a 2x2 matrix ( \begin{bmatrix} a & b \ c & d \end{bmatrix} ), the inverse is ( \frac{1}{\det(M)} \begin{bmatrix} d & -b \ -c & a \end{bmatrix} ).

We know ( \det(Dg(0,1)) = -2e ), so:
[
\left[ Dg(0,1) \right]^{-1} = \frac{1}{-2e} \begin{bmatrix} 0 & -2 \ -e & 4 \end{bmatrix} = \begin{bmatrix} 0 & \frac{1}{e} \ \frac{1}{2} & -\frac{2}{e} \end{bmatrix}
]

Step 3: Multiply the matrices #

Now calculate the product ( Df(0,1) \cdot \left[ Dg(0,1) \right]^{-1} ):

[
\begin{bmatrix} 3 & -2 \ 2 & 1 \ 1 & 3 \end{bmatrix} \begin{bmatrix} 0 & \frac{1}{e} \ \frac{1}{2} & -\frac{2}{e} \end{bmatrix} = \begin{bmatrix} 3(0) + (-2)\left(\frac{1}{2}\right) & 3\left(\frac{1}{e}\right) + (-2)\left(-\frac{2}{e}\right) \ 2(0) + 1\left(\frac{1}{2}\right) & 2\left(\frac{1}{e}\right) + 1\left(-\frac{2}{e}\right) \ 1(0) + 3\left(\frac{1}{2}\right) & 1\left(\frac{1}{e}\right) + 3\left(-\frac{2}{e}\right) \end{bmatrix}
]

Simplify each entry:

• First row: ( -1 ), ( \frac{3}{e} + \frac{4}{e} = \frac{7}{e} )
• Second row: ( \frac{1}{2} ), ( \frac{2}{e} - \frac{2}{e} = 0 )
• Third row: ( \frac{3}{2} ), ( \frac{1}{e} - \frac{6}{e} = -\frac{5}{e} )

So the final derivative is:
[
D(f \circ g^{-1})(2,0) = \begin{bmatrix} -1 & \frac{7}{e} \ \frac{1}{2} & 0 \ \frac{3}{2} & -\frac{5}{e} \end{bmatrix}
]

内容的提问来源于stack exchange，提问作者user424241