Math Problem Statement

Evaluate all second-order partial derivatives of \( f(x, y) = e^{x^2 + y^3} \) and use the chain rule to evaluate \( \frac{\partial w}{\partial x} \), where \( w = u^3 s - u s^2 \), \( u = x^2 + y \), and \( s = x^3 \).

Solution

Let's solve each part of the question step-by-step.

Part (a): Find all second-order partial derivatives of f(x,y)=ex2+y3f(x, y) = e^{x^2 + y^3}

  1. First, calculate the first-order partial derivatives of ff:

    • Partial derivative with respect to xx: fx=x(ex2+y3)=ex2+y3x(x2+y3)=ex2+y32x=2xex2+y3f_x = \frac{\partial}{\partial x} \left( e^{x^2 + y^3} \right) = e^{x^2 + y^3} \cdot \frac{\partial}{\partial x} (x^2 + y^3) = e^{x^2 + y^3} \cdot 2x = 2x e^{x^2 + y^3}

    • Partial derivative with respect to yy: fy=y(ex2+y3)=ex2+y3y(x2+y3)=ex2+y33y2=3y2ex2+y3f_y = \frac{\partial}{\partial y} \left( e^{x^2 + y^3} \right) = e^{x^2 + y^3} \cdot \frac{\partial}{\partial y} (x^2 + y^3) = e^{x^2 + y^3} \cdot 3y^2 = 3y^2 e^{x^2 + y^3}

  2. Now, calculate the second-order partial derivatives:

    • Second partial derivative with respect to xx ( fxxf_{xx} ): fxx=x(2xex2+y3)=2ex2+y3+2xex2+y32x=(2+4x2)ex2+y3f_{xx} = \frac{\partial}{\partial x} (2x e^{x^2 + y^3}) = 2 e^{x^2 + y^3} + 2x \cdot e^{x^2 + y^3} \cdot 2x = (2 + 4x^2) e^{x^2 + y^3}

    • Second partial derivative with respect to yy ( fyyf_{yy} ): fyy=y(3y2ex2+y3)=6yex2+y3+3y2ex2+y33y2=(6y+9y4)ex2+y3f_{yy} = \frac{\partial}{\partial y} (3y^2 e^{x^2 + y^3}) = 6y e^{x^2 + y^3} + 3y^2 \cdot e^{x^2 + y^3} \cdot 3y^2 = (6y + 9y^4) e^{x^2 + y^3}

    • Mixed partial derivative with respect to xx and yy ( fxy=fyxf_{xy} = f_{yx} ): fxy=y(2xex2+y3)=2xex2+y33y2=6xy2ex2+y3f_{xy} = \frac{\partial}{\partial y} (2x e^{x^2 + y^3}) = 2x \cdot e^{x^2 + y^3} \cdot 3y^2 = 6xy^2 e^{x^2 + y^3}

    So, the second-order partial derivatives are:

    • fxx=(2+4x2)ex2+y3f_{xx} = (2 + 4x^2) e^{x^2 + y^3}
    • fyy=(6y+9y4)ex2+y3f_{yy} = (6y + 9y^4) e^{x^2 + y^3}
    • fxy=fyx=6xy2ex2+y3f_{xy} = f_{yx} = 6xy^2 e^{x^2 + y^3}

Part (b): Use the chain rule to evaluate wx\frac{\partial w}{\partial x}, where w=u3sus2w = u^3 s - u s^2, u=x2+yu = x^2 + y, and s=x3s = x^3.

  1. Find the partial derivatives of ww with respect to uu and ss:

    • With respect to uu: wu=3u2ss2\frac{\partial w}{\partial u} = 3u^2 s - s^2

    • With respect to ss: ws=u32us\frac{\partial w}{\partial s} = u^3 - 2u s

  2. Now, find the partial derivatives of uu and ss with respect to xx:

    • For u=x2+yu = x^2 + y: ux=2x\frac{\partial u}{\partial x} = 2x

    • For s=x3s = x^3: sx=3x2\frac{\partial s}{\partial x} = 3x^2

  3. Apply the chain rule to find wx\frac{\partial w}{\partial x}: wx=wuux+wssx\frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial s} \cdot \frac{\partial s}{\partial x} Substituting the values we found: wx=(3u2ss2)(2x)+(u32us)(3x2)\frac{\partial w}{\partial x} = (3u^2 s - s^2)(2x) + (u^3 - 2u s)(3x^2)

  4. Express uu and ss in terms of xx and yy if needed:

    • u=x2+yu = x^2 + y
    • s=x3s = x^3

Thus, the final expression for wx\frac{\partial w}{\partial x} is: wx=(3(x2+y)2x3(x3)2)(2x)+((x2+y)32(x2+y)x3)(3x2)\frac{\partial w}{\partial x} = (3(x^2 + y)^2 \cdot x^3 - (x^3)^2)(2x) + ((x^2 + y)^3 - 2(x^2 + y) \cdot x^3)(3x^2)

Would you like further clarification on any part of this solution?

Further Questions for Practice:

  1. How would you find third-order partial derivatives of f(x,y)f(x, y)?
  2. Can you simplify the expression for wx\frac{\partial w}{\partial x} further?
  3. What would change in part (b) if u=x+y2u = x + y^2 instead?
  4. How does the chain rule differ when there are three variables involved?
  5. How can you verify the mixed partial derivatives are equal, i.e., fxy=fyxf_{xy} = f_{yx}?

Tip: Always double-check your substitution of terms when applying the chain rule to avoid common mistakes in calculations.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Chain Rule
Multivariable Calculus

Formulas

Second-order partial derivatives: \( f_{xx}, f_{yy}, f_{xy} \)
Chain rule for partial derivatives: \( \frac{\partial w}{\partial x} = \frac{\partial w}{\partial u} \cdot \frac{\partial u}{\partial x} + \frac{\partial w}{\partial s} \cdot \frac{\partial s}{\partial x} \)

Theorems

Mixed Partial Derivative Theorem (Schwarz's Theorem)

Suitable Grade Level

Undergraduate Calculus