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Question text 6Marks Consider the function [math: f(x,y)=x2y2+yex−1.] f(x,y) = x^2y^2 +y e^{x-1} . a) Calculate the following partial derivatives at the point [math: (1,1)]: [math: ∂f∂x=]\frac{\partial f}{\partial x}= Answer 1[input] [math: ∂f∂y=]\frac{\partial f}{\partial y}= Answer 2[input] [math: ∂2f∂x2=]\frac{\partial^2 f}{\partial x^2}= Answer 3[input] [math: ∂2f∂x∂y=]\frac{\partial^2 f}{\partial x\partial y}= Answer 4[input] [math: ∂2f∂y2=]\frac{\partial^2 f}{\partial y^2}= Answer 5[input] b) A tangent vector to the level set of [math: f] at [math: (1,1)] is [math: (1,] Answer 6[input][math: )].Notes Report question issue Question 5 Notes

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We start by restating what is given and then compute each requested derivative step by step so you can see how the values are obtained. Function: f(x, y) = x^2 y^2 + y e^{x-1}. a) Partial derivatives at (1, 1) - ∂f/∂x: Differentiate f with respect to x, treating y as constant. The term x^2 y^2 gives 2x y^2. The term y e^{x-1} gives y e^{x-1} when differentiating with respect to x. Thus, ∂f/∂x = 2x y^2 + y e^{x-1}. Evaluate at (1, 1): ∂f/∂x (1,1) = 2(1)(1)^2 + (1) e^{1-1} = 2 + 1 = 3. So the numeric value is 3. - ∂f/∂y: Differentiate f with respect to y, treating x as constant. The term x^2 y^2 di......Login to view full explanation

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