Question

The equation \[ xy - z \log y + e^{xz} = 1 \] can be solved in a neighborhood of the point \( (0, 1, 1) \) as \( y = f(x, z) \) for some continuously differentiable function \( f \). Then

• A. \( \nabla f(0, 1) = (2, 0) \)
• B. \( \nabla f(0, 1) = (0, 2) \)
• C. \( \nabla f(0, 1) = (0, 1) \)
• D. \( \nabla f(0, 1) = (1, 0) \)

Hint:

When solving implicit equations for one variable, differentiate the equation with respect to the variables and use the chain rule to compute the gradient.

Answer 1

The Correct Option is A

We are given the equation \( xy - z \log y + e^{xz} = 1 \) and asked to find the gradient of \( f(x, z) \) at the point \( (0, 1) \). To solve this, we differentiate the equation implicitly with respect to \( x \) and \( z \). At \( (0, 1, 1) \), the equation becomes: \[ xy - z \log y + e^{xz} = 1 \text{which simplifies to} 0 \times 1 - 1 \log 1 + e^{0} = 1. \] This is true, so the point \( (0, 1, 1) \) satisfies the equation. Now, differentiating with respect to \( x \) and \( z \), we obtain: \[ \frac{\partial}{\partial x} (xy - z \log y + e^{xz}) = y, \frac{\partial}{\partial z} (xy - z \log y + e^{xz}) = - \log y + x e^{xz}. \] At \( (0, 1) \), we compute these partial derivatives and find that \( \nabla f(0, 1) = (2, 0) \), corresponding to option (A).