Let f(x,y)=x2 −y2. find the gradient of f at the point (√2,1). sketch the level curve of f through this point, together with the gradient at that point. g

Answer:[tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2 \hat{\j}[/tex]General Formulas and Concepts:
CalculusDifferentiationDerivativesDerivative NotationDerivative Rule [Basic Power Rule]:f(x) = cxⁿf’(x) = c·nxⁿ⁻¹Multivariable CalculusDifferentiationPartial DerivativesDerivative NotationGradient:                                                                                                               [tex]\displaystyle \nabla f(x, y, z) = \frac{\partial f}{\partial x} \hat{\i} + \frac{\partial f}{\partial y} \hat{\j} + \frac{\partial f}{\partial z} \hat{\text{k}}[/tex]Gradient Property [Addition/Subtraction]:                                                           [tex]\displaystyle \nabla \big[ f(x) + g(x) \big] = \nabla f(x) + \nabla g(x)[/tex]Step-by-step explanation:Step 1: DefineIdentify.[tex]\displaystyle f(x, y) = x^2 - y^2[/tex][tex]\displaystyle P(\sqrt{2}, 1)[/tex]Step 2: Find Gradient[Function] Differentiate [Gradient]:                                                              [tex]\displaystyle \nabla f(x, y) = \frac{\partial f}{\partial x} \bigg[ x^2 - y^2 \bigg] \hat{\i} + \frac{\partial f}{\partial y} \bigg[ x^2 - y^2 \bigg] \hat{\j}[/tex][Gradient] Rewrite [Gradient Property - Addition/Subtraction]:                [tex]\displaystyle \nabla f(x, y) = \bigg[ \frac{\partial f}{\partial x}(x^2) - \frac{\partial f}{\partial x}(y^2) \bigg] \hat{\i} + \bigg[ \frac{\partial f}{\partial y}(x^2) - \frac{\partial f}{\partial y}(y^2) \bigg] \hat{\j}[/tex][Gradient] Differentiate [Derivative Rule - Basic Power Rule]:                  [tex]\displaystyle \nabla f(x, y) = 2x \hat{\i} - 2y \hat{\j}[/tex][Gradient] Substitute in point:                                                                     [tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2(1) \hat{\j}[/tex][Gradient] Simplify:                                                                                       [tex]\displaystyle \nabla f(\sqrt{2}, 1) = 2\sqrt{2} \hat{\i} - 2 \hat{\j}[/tex]∴ the gradient of the given f(x, y) function is equal to <2√2, -2>.---Learn more about gradient: more about multivariable calculus: : Multivariable CalculusUnit: Directional Derivatives