Math Problem Statement

Find all second partial derivatives of f(x, y) = e xy sin x

Solution

To find the second partial derivatives of the function f(x,y)=exysinxf(x, y) = e^{xy} \sin x, we proceed by computing the first and then the second partial derivatives with respect to xx and yy.

Step 1: Compute the First Partial Derivatives

  1. Partial derivative with respect to xx: fx=x(exysinx)f_x = \frac{\partial}{\partial x}(e^{xy} \sin x) Using the product rule and the chain rule, we get: fx=exycosx+yexysinxf_x = e^{xy} \cos x + y e^{xy} \sin x

  2. Partial derivative with respect to yy: fy=y(exysinx)f_y = \frac{\partial}{\partial y}(e^{xy} \sin x) Differentiating with respect to yy, treating xx as a constant: fy=xexysinxf_y = x e^{xy} \sin x

Step 2: Compute the Second Partial Derivatives

  1. Second partial derivative with respect to xx (i.e., fxxf_{xx}): fxx=x(fx)=x(exycosx+yexysinx)f_{xx} = \frac{\partial}{\partial x}(f_x) = \frac{\partial}{\partial x}\left(e^{xy} \cos x + y e^{xy} \sin x\right) Applying the product rule to each term: fxx=exy(sinx)+yexycosx+y(exycosx+yexysinx)f_{xx} = e^{xy}(-\sin x) + y e^{xy} \cos x + y \left(e^{xy} \cos x + y e^{xy} \sin x\right) Simplifying: fxx=exy(sinx+ycosx)+yexycosx+y2exysinxf_{xx} = e^{xy}(-\sin x + y \cos x) + y e^{xy} \cos x + y^2 e^{xy} \sin x

  2. Mixed partial derivative with respect to xx and yy (i.e., fxyf_{xy}): fxy=y(fx)=y(exycosx+yexysinx)f_{xy} = \frac{\partial}{\partial y}(f_x) = \frac{\partial}{\partial y}\left(e^{xy} \cos x + y e^{xy} \sin x\right) Differentiating each term with respect to yy: fxy=xexycosx+exysinx+yxexysinxf_{xy} = x e^{xy} \cos x + e^{xy} \sin x + y x e^{xy} \sin x Simplifying: fxy=exy(xcosx+sinx+xysinx)f_{xy} = e^{xy}(x \cos x + \sin x + xy \sin x)

  3. Second partial derivative with respect to yy (i.e., fyyf_{yy}): fyy=y(fy)=y(xexysinx)f_{yy} = \frac{\partial}{\partial y}(f_y) = \frac{\partial}{\partial y}(x e^{xy} \sin x) Using the product rule: fyy=x2exysinxf_{yy} = x^2 e^{xy} \sin x

Summary of Second Partial Derivatives

fxx=exy(sinx+ycosx)+yexycosx+y2exysinxf_{xx} = e^{xy}(-\sin x + y \cos x) + y e^{xy} \cos x + y^2 e^{xy} \sin x fxy=exy(xcosx+sinx+xysinx)f_{xy} = e^{xy}(x \cos x + \sin x + xy \sin x) fyy=x2exysinxf_{yy} = x^2 e^{xy} \sin x

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Related Questions

  1. What are higher-order partial derivatives for a function of three variables?
  2. How is the mixed partial derivative theorem (Clairaut's theorem) applied in this context?
  3. How can partial derivatives be visualized on a surface?
  4. How does the product rule extend to functions with exponential and trigonometric components?
  5. What are applications of mixed partial derivatives in physics and engineering?

Tip

When differentiating products involving exponential and trigonometric functions, apply both the product and chain rules carefully to each term.

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Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Second Partial Derivatives
Multivariable Calculus
Product Rule
Chain Rule

Formulas

First partial derivative: f_x = ∂/∂x (e^(xy) sin x), f_y = ∂/∂y (e^(xy) sin x)
Second partial derivatives: f_xx, f_xy, f_yy

Theorems

Mixed Partial Derivative Theorem (Clairaut's Theorem)

Suitable Grade Level

Undergraduate Calculus