Great question! Let's break it down step by step. ### Given: - \( f : \mathbb{R}^2 \to \mathbb{R} \) - \( g(x, y) = f(x+y, x-y) \) - \( Df \) and \( Dg \) are the derivative (Jacobian) matrices of \( f \) and \( g \) respectively. ### The question: Is it **always** true that \[ (Dg)(x, y) = (Df)(x+y, x-y)? \] ### Let's analyze #### 1. **Write out what \( Dg \) is:** By the chain rule, for \( g(x, y) = f(u(x, y), v(x, y)) \) where: - \( u(x, y) = x+y \) - \( v(x, y) = x-y \) We have: \[ Dg(x, y) = Df(u, v) \cdot D(u, v) \] where \( Df(u, v) \) is the row vector of partial derivatives of \( f \) at \( (u, v) \), and \( D(u, v) \) is the Jacobian matrix of \( (u, v) \) with respect to \( (x, y) \). #### 2. **Compute these matrices** - \( Df(u, v) = \left[ \frac{\partial f}{\partial u}, \frac{\partial f}{\partial v} \right] \) evaluated at \( (u, v) = (x+y, x-y) \) - \( D(u, v) = \begin{bmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{bmatrix} = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \) But since \( f \) is scalar-valued, the derivative \( Df \) is a **row vector** (or gradient). For \( g \), we have: \[ Dg(x, y) = \nabla f(u, v) \cdot D(u, v) \] But more commonly, for scalar-valued functions, the gradient of \( g \) is: \[ \nabla g(x, y) = \left[ \frac{\partial g}{\partial x}, \frac{\partial g}{\partial y} \right] \] By the multivariable chain rule: \[ \frac{\partial g}{\partial x} = \frac{\partial f}{\partial u} \frac{\partial u}{\partial x} + \frac{\partial f}{\partial v} \frac{\partial v}{\partial x} = f_u(x+y, x-y)(1) + f_v(x+y, x-y)(1) \] \[ \frac{\partial g}{\partial y} = f_u(x+y, x-y)(1) + f_v(x+y, x-y)(-1) \] So, \[ \nabla g(x, y) = \left[ f_u(x+y, x-y) + f_v(x+y, x-y),\quad f_u(x+y, x-y) - f_v(x+y, x-y) \right] \] #### 3. **What is \( Df(x+y, x-y) \)?** This is just the row vector: \[ \left[ f_u(x+y, x-y),\ f_v(x+y, x-y) \right] \] #### 4. **Are they equal?** No! **\( Dg(x, y) \) is not simply \( Df(x+y, x-y) \).** - \( Df(x+y, x-y) \) is a row vector with two components. - \( Dg(x, y) \) is also a row vector with two components, but the components are linear combinations of those of \( Df \), as shown above. So, **in general,** \[ (Dg)(x, y) \neq (Df)(x+y, x-y) \] ### **Why?** Because the chain rule introduces a mixing (linear transformation) of the partial derivatives, due to the way \( (x, y) \) map into the arguments of \( f \). --- ## **Conclusion** **You are correct:** \[ (Dg)(x, y) \neq (Df)(x+y, x-y) \quad \text{in general.} \] The correct relationship is: \[ (Dg)(x, y) = (Df)(x+y, x-y) \cdot \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix} \] (for row vectors, matrix multiplication; for gradients, it's a linear combination as above). **So the statement is NOT always true!** Let me know if you want a concrete numerical example!