Questions
Questions
Single choice

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}

Options
A.𝑦 𝑥 2 + 6 𝑦 + 𝑥 2 + 6 𝑦
B.3 𝑦 𝑥 2 + 6 𝑦
C.3 𝑦 𝑥 2 + 6 𝑦 + 𝑥 2 + 6 𝑦
D.( 2 𝑥 + 6 ) 𝑦 𝑥 2 + 6 𝑦
E.𝑥 2 + 6 𝑦
View Explanation

View Explanation

Verified Answer
Please login to view
Step-by-Step Analysis
We begin by clearly restating what is being asked: a question about the derivative ∂z/∂y for a relationship involving z, y, and x, where z is defined implicitly via z = yx^2 + 6yz = y√(x^2 + 6y). Now we evaluate each answer choice in turn to see which expression for ∂z/∂y aligns with the given relation. Option 1: y x^2 + 6 y + x^2 + 6 y - This option combines two straightforward terms, x^2 y and 6y, with an extra x^2 and another 6y. The appearance of an isolated x^2 term (without y) suggests it does not consistently reflect a derivative that should depend on z or on the structure of the original equation. In particular, unless there is a reason for x^2 to appear unmultiplied by y or z in the derivative, this form seems unlikely to be th......Login to view full explanation

Log in for full answers

We've collected over 50,000 authentic exam questions and detailed explanations from around the globe. Log in now and get instant access to the answers!

Similar Questions

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

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

More Practical Tools for Students Powered by AI Study Helper

Join us and instantly unlock extensive past papers & exclusive solutions to get a head start on your studies!