题目
题目
多项填空题

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

题目图片
查看解析

查看解析

标准答案
Please login to view
思路分析
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

登录即可查看完整答案

我们收录了全球超50000道考试原题与详细解析,现在登录,立即获得答案。

类似问题

Question at position 2 If z=yx2+6yz=y\sqrt{x^2+6y}, then ∂z∂y=\frac{\partial z}{\partial y\:}= yx2+6y+x2+6y\frac{y}{\sqrt{x^2+6y}}+\sqrt{x^2+6y}3yx2+6y\frac{3y}{\sqrt{x^2+6y}}3yx2+6y+x2+6y\frac{3y}{\sqrt{x^2+6y}}+\sqrt{x^2+6y}(2x+6)yx2+6y\left(2x+6\right)y\sqrt{x^2+6y}x2+6y\sqrt{x^2+6y}

Question at position 1 If f(x,y,z)=x2yz2+xy2z+xyf\left(x,y,z\right)=x^2yz^2+xy^2z+xy, then fx(1, 2, 3) =36.none of the above55.50.48.

Question textLet [math: f(x,y)=ey+x2]f(x,y)={e^{y+x^2}}. What is [math: ∂f∂x(0,0)+∂f∂y(0,0)]\frac{\partial f}{\partial x}(0,0) + \frac{\partial f}{\partial y}(0,0), the value of [math: ∂f∂x+∂f∂y]\frac{\partial f}{\partial x} + \frac{\partial f}{\partial y} at the point [math: (0,0)] ?[input] Check Question 67

Question textLet [math: f(x,y)=ycos⁡(xy)]f(x,y) = {y\,\cos \left( x\,y \right)}. The partial derivative [math: ∂f∂x]\frac{\partial f}{\partial {x}} is[input] Your last answer was interpreted as follows: rickli12128@outlook.comThis answer is invalid. Expected "!!", "!", "#", "#pm#", "%and", "%or", "(", "*", "**", "+", "+-", ",", "-", ".", "/", ":", "::", "::=", ":=", "<", "<=", "=", ">", ">=", "@@IS@@", "@@Is@@", "[", "^", "^^", "and", "implies", "nand", "nor", "nounand", "nounor", "or", "xnor", "xor", "~", [;$], end of input or whitespace but "@" found.Your answer should contain the variables [math: x] and [math: y].Check Question 66

更多留学生实用工具

加入我们，立即解锁 海量真题 与 独家解析，让复习快人一步！