Let's solve **part (a)** step by step as requested. We will use the **Inverse Function Theorem**, which states that if the Jacobian determinant at a point is nonzero, then a local differentiable inverse exists. --- ### **a) \( f(u,v) = (uv, 3u - 2v) \) at any \((a, b) \in \mathbb{R}^2\)** #### **Step 1: Compute the Jacobian and check invertibility at \((a,b)\)** The Jacobian matrix \(Df(u,v)\) is: \[ Df(u,v) = \begin{bmatrix} \frac{\partial}{\partial u}(uv) & \frac{\partial}{\partial v}(uv) \\ \frac{\partial}{\partial u}(3u-2v) & \frac{\partial}{\partial v}(3u-2v) \end{bmatrix} = \begin{bmatrix} v & u \\ 3 & -2 \end{bmatrix} \] The determinant is: \[ \det Df(u,v) = (v)(-2) - (u)(3) = -2v - 3u \] **Explanation:** The Inverse Function Theorem applies if the Jacobian determinant is nonzero. So, as long as \(-2v - 3u \neq 0\), the local inverse exists and is differentiable. --- #### **Step 2: The inverse function exists and is differentiable** **Explanation:** For any \((a, b)\) such that \(-2b - 3a \neq 0\), \(f^{-1}\) exists and is differentiable in some open set containing \((a, b)\). Thus, for all points in \(\mathbb{R}^2\) except those on the line \(2v + 3u = 0\). --- #### **Step 3: Compute \( D(f^{-1})(a, b) \)** By the Inverse Function Theorem: \[ D(f^{-1})(f(a,b)) = [Df(a,b)]^{-1} \] First, compute the inverse of the Jacobian at \((a, b)\): \[ Df(a,b) = \begin{bmatrix} b & a \\ 3 & -2 \end{bmatrix} \] Its determinant is \(-2b - 3a\). The inverse is: \[ [Df(a,b)]^{-1} = \frac{1}{-2b - 3a} \begin{bmatrix} -2 & -a \\ -3 & b \end{bmatrix} \] **Explanation:** This gives the derivative matrix for the inverse function at \((a, b)\). --- ### **Final Answer for (a):** - \(f^{-1}\) exists and is differentiable in some open set containing \((a, b)\) **if and only if** \(2b + 3a \neq 0\). - The derivative matrix of the inverse at \((a, b)\) is: \[ D(f^{-1})(a, b) = \frac{1}{-2b - 3a} \begin{bmatrix} -2 & -a \\ -3 & b \end{bmatrix} \] --- **Would you like to proceed similarly for part (b)?**

Let's proceed step-by-step for each part (b), (c), and (d), following the same approach: compute the Jacobian, check invertibility, and find the derivative of the inverse where applicable. --- ### **b) \(f(u,v) = (u - v, \cos u \sin v)\) at \((a,b) = (0,1)\)** --- ### **Step 1: Compute the Jacobian matrix \(Df(u,v)\)** \[ Df(u,v) = \begin{bmatrix} \frac{\partial}{\partial u}(u - v) & \frac{\partial}{\partial v}(u - v) \\ \frac{\partial}{\partial u}(\cos u \sin v) & \frac{\partial}{\partial v}(\cos u \sin v) \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -\sin u \sin v & \cos u \cos v \end{bmatrix} \] Evaluate at \((a, b) = (0, 1)\): \[ Df(0,1) = \begin{bmatrix} 1 & -1 \\ -\sin 0 \cdot \sin 1 & \cos 0 \cdot \cos 1 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & \cos 1 \end{bmatrix} \] --- ### **Step 2: Check invertibility at \((0,1)\)** The determinant: \[ \det Df(0,1) = (1)(\cos 1) - (-1)(0) = \cos 1 \neq 0 \] Since \(\cos 1 \neq 0\), the Jacobian is invertible at \((0, 1)\). **Explanation:** Therefore, by the Inverse Function Theorem, \(f^{-1}\) exists and is differentiable in some open neighborhood of \(f(0,1)\). --- ### **Step 3: Compute \(D(f^{-1})(0,1)\)** \[ D(f^{-1})(f(0,1)) = [Df(0,1)]^{-1} \] Inverse of the Jacobian: \[ [Df(0,1)]^{-1} = \frac{1}{\cos 1} \begin{bmatrix} \cos 1 & 1 \\ 0 & 1 \end{bmatrix} \] (Note: For a 2x2 matrix \(\begin{bmatrix}a & b \\ c & d\end{bmatrix}\), the inverse is \(\frac{1}{ad - bc} \begin{bmatrix}d & -b \\ -c & a \end{bmatrix}\). Applying this to our matrix: \[ \begin{bmatrix} 1 & -1 \\ 0 & \cos 1 \end{bmatrix} \] determinant: \(\cos 1\). The inverse: \[ \frac{1}{\cos 1} \begin{bmatrix} \cos 1 & 1 \\ 0 & 1 \end{bmatrix} \] which simplifies to: \[ \begin{bmatrix} 1 & \frac{1}{\cos 1} \\ 0 & \frac{1}{\cos 1} \end{bmatrix} \] **Final:** \[ D(f^{-1})(f(0,1)) = \begin{bmatrix} 1 & \frac{1}{\cos 1} \\ 0 & \frac{1}{\cos 1} \end{bmatrix} \] --- ## **Summary for (b):** - The inverse exists and is differentiable near \(f(0,1)\) because the Jacobian determinant is \(\cos 1 \neq 0\). - The derivative of the inverse at that point is: \[ D(f^{-1})(f(0,1)) = \begin{bmatrix} 1 & \frac{1}{\cos 1} \\ 0 & \frac{1}{\cos 1} \end{bmatrix} \] --- ### **c) \(f(u,v) = (u v^2, u^2 v^2)\) at \((12, 13)\)** --- ### **Step 1: Compute the Jacobian \(Df(u,v)\)** \[ Df(u,v) = \begin{bmatrix} \frac{\partial}{\partial u}(u v^2) & \frac{\partial}{\partial v}(u v^2) \\ \frac{\partial}{\partial u}(u^2 v^2) & \frac{\partial}{\partial v}(u^2 v^2) \end{bmatrix} = \begin{bmatrix} v^2 & 2 u v \\ 2 u v^2 & 2 u^2 v \end{bmatrix} \] Evaluate at \((12, 13)\): \[ Df(12,13) = \begin{bmatrix} (13)^2 & 2 \cdot 12 \cdot 13 \\ 2 \cdot 12 \cdot 13^2 & 2 \cdot 12^2 \cdot 13 \end{bmatrix} \] Calculate each element: - \(v^2 = 13^2 = 169\) - \(2 u v = 2 \times 12 \times 13 = 312\) - \(2 u v^2 = 2 \times 12 \times 169 = 2 \times 12 \times 169 = 2 \times 2028 = 4056\) - \(2 u^2 v = 2 \times 12^2 \times 13 = 2 \times 144 \times 13 = 2 \times 1872 = 3744\) So, \[ Df(12,13) = \begin{bmatrix} 169 & 312 \\ 4056 & 3744 \end{bmatrix} \] --- ### **Step 2: Check invertibility** Calculate the determinant: \[ \det Df(12,13) = 169 \times 3744 - 312 \times 4056 \] Compute: - \(169 \times 3744\): \[ 169 \times 3744 = (169 \times 3000) + (169 \times 744) = 507000 + (169 \times 744) \] Calculate \(169 \times 744\): \[ 169 \times 744 = (169 \times 700) + (169 \times 44) = 118300 + 7436 = 125736 \] Sum: \[ 507000 + 125736 = 632736 \] Now, compute \(312 \times 4056\): \[ 312 \times 4056 = (300 \times 4056) + (12 \times 4056) = 1,216,800 + 48,672 = 1,265,472 \] Determinant: \[ 632,736 - 1,265,472 = -632,736 \neq 0 \] Thus, the Jacobian is invertible at \((12, 13)\). **Explanation:** The inverse exists and is differentiable near the point. --- ### **Step 3: Compute the inverse of the Jacobian** \[ [Df(12,13)]^{-1} = \frac{1}{\det} \begin{bmatrix} 3744 & -312 \\ -4056 & 169 \end{bmatrix} \] where the matrix of cofactors is: \[ \begin{bmatrix} 3744 & -312 \\ -4056 & 169 \end{bmatrix} \] and the determinant is \(-632,736\). **Final:** \[ D(f^{-1})(f(12,13)) = \frac{1}{-632736} \begin{bmatrix} 3744 & -312 \\ -4056 & 169 \end{bmatrix} \] --- ## **Summary for (c):** - The inverse exists and is differentiable near \((12, 13)\) because the Jacobian determinant is non-zero. - The derivative of the inverse at that point is: \[ D(f^{-1})(f(12,13)) = \frac{1}{-632736} \begin{bmatrix} 3744 & -312 \\ -4056 & 169 \end{bmatrix} \] --- ### **d) \(f(u,v) = (u e^u, \log v - \cos u)\) at \((0, -1)\)** --- ### **Step 1: Compute the Jacobian \(Df(u,v)\)** \[ Df(u,v) = \begin{bmatrix} \frac{\partial}{\partial u}(u e^u) & \frac{\partial}{\partial v}(u e^u) \\ \frac{\partial}{\partial u}(\log v - \cos u) & \frac{\partial}{\partial v}(\log v - \cos u) \end{bmatrix} \] Calculate derivatives: - \(\frac{\partial}{\partial u}(u e^u) = e^u + u e^u = e^u (1 + u)\) - \(\frac{\partial}{\partial v}(u e^u) = 0\) - \(\frac{\partial}{\partial u}(\log v - \cos u) = \sin u\) (since derivative of \(- \cos u\) is \(\sin u\)) - \(\frac{\partial}{\partial v}(\log v - \cos u) = \frac{1}{v}\) Evaluate at \((0, -1)\): \[ Df(0, -1) = \begin{bmatrix} e^{0}(1 + 0) & 0 \\ \sin 0 & \frac{1}{-1} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \] --- ### **Step 2: Check invertibility** Determinant: \[ \det Df(0, -1) = (1)(-1) - (0)(0) = -1 \neq 0 \] **Explanation:** The Jacobian is invertible at \((0, -1)\). --- ### **Step 3: Compute the inverse** \[ [Df(0, -1)]^{-1} = \frac{1}{-1} \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \] **Final:** \[ D(f^{-1})(f(0, -1)) = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} \] --- ## **Final Summary:** | Part | Conditions for Invertibility | Derivative of \(f^{-1}\) at \((a,b)\) | |---|---|---| | (a) | \(-2b - 3a \neq 0\) | \(\displaystyle \frac{1}{-2b - 3a} \begin{bmatrix} -2 & -a \\ -3 & b \end{bmatrix}\) | | (b) | \(\cos 1 \neq 0\) | \(\displaystyle \begin{bmatrix} 1 & \frac{1}{\cos 1} \\ 0 & \frac{1}{\cos 1} \end{bmatrix}\) | | (c) | determinant \(\neq 0\) (as calculated) | inverse matrix from above divided by \(\det\) | | (d) | \(\det \neq 0\) (which is \(-1\)) | \(\begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\) | --- Would you like further explanations or calculations?