Math Problem Statement

Solution

To solve the given problem of finding the total differential for each function, we will need to differentiate each function with respect to its variables. The total differential dzdz of a function z=f(x,y)z = f(x, y) is given by:

dz=zxdx+zydydz = \frac{\partial z}{\partial x} dx + \frac{\partial z}{\partial y} dy

Let's find the total differentials for each function provided in the image.

  1. z=xy+xyz = xy + \frac{x}{y}

    To find dzdz:

    zx=y+1y\frac{\partial z}{\partial x} = y + \frac{1}{y} zy=xxy2\frac{\partial z}{\partial y} = x - \frac{x}{y^2}

    Therefore, the total differential is:

    dz=(y+1y)dx+(xxy2)dydz = \left( y + \frac{1}{y} \right) dx + \left( x - \frac{x}{y^2} \right) dy

  2. z=eyxz = e^{\frac{y}{x}}

    To find dzdz:

    Let u=yxu = \frac{y}{x}, then z=euz = e^u.

    zx=zuux=eu(yx2)=yx2eyx\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} = e^u \cdot \left( -\frac{y}{x^2} \right) = -\frac{y}{x^2} e^{\frac{y}{x}}

    zy=zuuy=eu1x=1xeyx\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} = e^u \cdot \frac{1}{x} = \frac{1}{x} e^{\frac{y}{x}}

    Therefore, the total differential is:

    dz=yx2eyxdx+1xeyxdydz = -\frac{y}{x^2} e^{\frac{y}{x}} dx + \frac{1}{x} e^{\frac{y}{x}} dy

  3. z=yx2+y2z = \frac{y}{\sqrt{x^2 + y^2}}

    To find dzdz:

    Let u=x2+y2u = x^2 + y^2, then z=yuz = \frac{y}{\sqrt{u}}.

    zx=zuux=(12yu3/2)2x=xy(x2+y2)3/2\frac{\partial z}{\partial x} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial x} = \left( -\frac{1}{2} \cdot \frac{y}{u^{3/2}} \right) \cdot 2x = -\frac{xy}{(x^2 + y^2)^{3/2}}

    zy=zuuy=(12yu3/2)2y+1u=y2(x2+y2)3/2+1x2+y2\frac{\partial z}{\partial y} = \frac{\partial z}{\partial u} \cdot \frac{\partial u}{\partial y} = \left( -\frac{1}{2} \cdot \frac{y}{u^{3/2}} \right) \cdot 2y + \frac{1}{\sqrt{u}} = -\frac{y^2}{(x^2 + y^2)^{3/2}} + \frac{1}{\sqrt{x^2 + y^2}}

    Simplifying the second term:

    zy=y2(x2+y2)3/2+x2+y2(x2+y2)3/2=x2(x2+y2)3/2\frac{\partial z}{\partial y} = -\frac{y^2}{(x^2 + y^2)^{3/2}} + \frac{x^2 + y^2}{(x^2 + y^2)^{3/2}} = \frac{x^2}{(x^2 + y^2)^{3/2}}

    Therefore, the total differential is:

    dz=xy(x2+y2)3/2dx+x2(x2+y2)3/2dydz = -\frac{xy}{(x^2 + y^2)^{3/2}} dx + \frac{x^2}{(x^2 + y^2)^{3/2}} dy

Would you like any more details on any of these differentials or have any questions? Here are five related questions you might consider:

  1. How do you find the partial derivative of a function?
  2. What is the geometric interpretation of the total differential?
  3. How do you apply the chain rule in multivariable calculus?
  4. Can you explain the significance of the total differential in physics?
  5. How do you handle differentials when working with implicit functions?

Tip: When calculating partial derivatives, ensure you treat all other variables as constants and only differentiate with respect to the variable of interest.

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Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Partial Derivatives
Chain Rule

Formulas

Total differential formula: dz = (∂z/∂x) dx + (∂z/∂y) dy
Partial derivatives for functions of multiple variables

Theorems

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Suitable Grade Level

Undergraduate