题目
题目

ENG1005 - MUM S2 2025 [FINAL REVISION] Quizzes

多项填空题

Question textLet [math: f(x,y)=ysin⁡(xy)]f(x,y) = {y\,\sin \left( x\,y \right)}. The partial derivative [math: ∂f∂y]\frac{\partial f}{\partial {y}} is[input] Your answer should contain the variables [math: x] and [math: y].Check Question 66

题目图片
查看解析

查看解析

标准答案
Please login to view
思路分析
The function given is f(x, y) = y sin(xy). To differentiate with respect to y, we use the product rule since f is a product of y and sin(x......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 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

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

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！