Math Problem Statement

provide brief and detailed solution as well as the final answer

Provide brief and detailed solution as well as the final answer for the following problems: 1. f(x,y) = x^2 - 2xy + 3y^2 2. f(x,y) = xe^{-y} + ye^{-2x} 3. z = 1/ln(x^2 + y^2) 4. u = e^{x/z} + e^{y/z} + e^{z/x} 5. z = e^{-y}cos(x-y) 6. z = e^ycosx + e^xcosy 7. z = sin^{-1}(x/y) 8. u = (x^3 - y^3)^2

Solution

Sure! Here's a brief explanation and solution for each problem:

4. f(x,y)=x22xy+3y2f(x,y) = x^2 - 2xy + 3y^2

  • Partial derivative with respect to xx: fx=x(x22xy+3y2)=2x2yf_x = \frac{\partial}{\partial x}(x^2 - 2xy + 3y^2) = 2x - 2y

  • Partial derivative with respect to yy: fy=y(x22xy+3y2)=2x+6yf_y = \frac{\partial}{\partial y}(x^2 - 2xy + 3y^2) = -2x + 6y

5. f(x,y)=xey+ye2xf(x,y) = xe^{-y} + ye^{-2x}

  • Partial derivative with respect to xx: fx=x(xey+ye2x)=ey2ye2xf_x = \frac{\partial}{\partial x}(xe^{-y} + ye^{-2x}) = e^{-y} - 2ye^{-2x}

  • Partial derivative with respect to yy: fy=y(xey+ye2x)=xey+e2xf_y = \frac{\partial}{\partial y}(xe^{-y} + ye^{-2x}) = -xe^{-y} + e^{-2x}

6. z=1ln(x2+y2)z = \frac{1}{\ln(x^2 + y^2)}

  • Partial derivative with respect to xx: zx=x(1ln(x2+y2))=2x(x2+y2)ln2(x2+y2)z_x = \frac{\partial}{\partial x}\left(\frac{1}{\ln(x^2 + y^2)}\right) = -\frac{2x}{(x^2 + y^2)\ln^2(x^2 + y^2)}

  • Partial derivative with respect to yy: zy=y(1ln(x2+y2))=2y(x2+y2)ln2(x2+y2)z_y = \frac{\partial}{\partial y}\left(\frac{1}{\ln(x^2 + y^2)}\right) = -\frac{2y}{(x^2 + y^2)\ln^2(x^2 + y^2)}

7. u=ex/z+ey/z+ez/xu = e^{x/z} + e^{y/z} + e^{z/x}

  • Show xux+yuy+zuz=0xu_x + yu_y + zu_z = 0:
    • Apply Euler’s homogeneous function theorem, assuming uu is homogeneous of degree zero.

8. z=eycos(xy)z = e^{-y}\cos(x-y)

  • Partial derivative with respect to xx: zx=eysin(xy)z_x = -e^{-y}\sin(x-y)

  • Partial derivative with respect to yy: zy=eycos(xy)+eysin(xy)z_y = -e^{-y}\cos(x-y) + e^{-y}\sin(x-y)

  • Show that zx+zy+z=0z_x + z_y + z = 0:

9. z=eycosx+excosyz = e^y\cos x + e^x\cos y

  • Second partial derivative with respect to xx: 2zx2=eycosx+excosy\frac{\partial^2 z}{\partial x^2} = -e^y\cos x + e^x\cos y

  • Second partial derivative with respect to yy: 2zy2=eycosxexcosy\frac{\partial^2 z}{\partial y^2} = e^y\cos x - e^x\cos y

10. z=sin1(xy)z = \sin^{-1}\left(\frac{x}{y}\right)

  • Apply implicit differentiation method:
  • Second partial derivatives can be complex but typically follow implicit differentiation rules.

11. u=(x3y3)2u = (x^3 - y^3)^2

  • Mixed partial derivative 3ux2y\frac{\partial^3 u}{\partial x^2 \partial y}:

  • First, find the first and second partial derivative with respect to xx, then with respect to yy.

  • Mixed partial derivative 3uy2x\frac{\partial^3 u}{\partial y^2 \partial x}:

  • Similarly, find the derivatives in the indicated order using chain and power rules.

These partial derivatives and evaluations involve applying basic differentiation rules, such as the product, quotient, and chain rules.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Euler’s Homogeneous Function Theorem
Implicit Differentiation
Second Partial Derivatives
Chain Rule
Power Rule

Formulas

f_x = ∂/∂x (f(x,y))
f_y = ∂/∂y (f(x,y))
Euler’s Homogeneous Function Theorem
Second Partial Derivative
Implicit Differentiation

Theorems

Euler’s Homogeneous Function Theorem

Suitable Grade Level

College/University Level