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
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
Show 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.
Updated On: Jan 7, 2026
- \( \nabla f(0, 1) = (2, 0) \)
- \( \nabla f(0, 1) = (0, 2) \)
- \( \nabla f(0, 1) = (0, 1) \)
- \( \nabla f(0, 1) = (1, 0) \)
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The Correct Option is A
Solution and Explanation
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).
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