Question

Find all values of \(x\) and \(y\) such that \(f_{x}(x, y)=0\) and \(f_{y}(x, y)=0\) simultaneously. $$ f(x, y)=\ln \left(x^{2}+y^{2}+1\right) $$

Short answer

The only values for \(x\) and \(y\) that satisfy \(f_x(x, y) = 0\) and \(f_y(x, y) = 0\) simultaneously are \(x = 0\) and \(y = 0\).

Step-by-step solution

Compute the partial derivatives

The first-order partial derivatives of the function \(f(x, y)\) are computed as follows: \(f_x = \frac{2x}{x^2 + y^2 + 1}\) and \(f_y = \frac{2y}{x^2 + y^2 + 1}\). Here, the chain rule is used to differentiate the inner function \(x^2 + y^2 + 1\) with respect to \(x\) and \(y\) respectively.

Equate the derivatives to zero

The critical points occur when both partial derivatives are zero. Therefore, we have two equations: \(f_x = 0\) or \(2x=0\) and \(f_y = 0\) or \(2y=0\). Hence, \(x = 0\) and \(y = 0\) are solutions from these equations.

Check if points solve both equations

Lastly, check that the proposed solution \(x=0\) and \(y=0\) simultaneously solve the two original equations \(f_x = 0\) and \(f_y = 0\). Substituting \(x = 0\) and \(y = 0\) into the equations, it is found that both are satisfied, validating the solution.