Math Problem Statement

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

Solution

To find the second partial derivatives of f(x,y)=exysinxf(x, y) = e^{xy} \sin x, we'll start by calculating the first partial derivatives with respect to xx and yy, then differentiate these results to obtain the second partial derivatives. Specifically, we'll focus on finding fxyf_{xy} (the mixed partial derivative).

Step 1: Compute the First Partial Derivatives

First Partial Derivative with Respect to xx:

f(x,y)=exysinxf(x, y) = e^{xy} \sin x Using the product rule, since f(x,y)=g(x,y)h(x)f(x, y) = g(x, y) \cdot h(x) where g(x,y)=exyg(x, y) = e^{xy} and h(x)=sinxh(x) = \sin x: fx=gxh+ghxf_x = g_x \cdot h + g \cdot h_x

  1. First, differentiate g(x,y)=exyg(x, y) = e^{xy} with respect to xx: gx=yexyg_x = y e^{xy}
  2. Differentiate h(x)=sinxh(x) = \sin x with respect to xx: hx=cosxh_x = \cos x Therefore, fx=(yexy)sinx+exycosxf_x = (y e^{xy}) \sin x + e^{xy} \cos x

First Partial Derivative with Respect to yy:

Now, differentiate f(x,y)=exysinxf(x, y) = e^{xy} \sin x with respect to yy: fy=y(exysinx)f_y = \frac{\partial}{\partial y} (e^{xy} \sin x) Since sinx\sin x is treated as a constant with respect to yy: fy=xexysinxf_y = x e^{xy} \sin x

Step 2: Compute the Second Partial Derivatives

Mixed Partial Derivative fxyf_{xy}:

To find fxyf_{xy}, we differentiate fxf_x with respect to yy: fx=(yexy)sinx+exycosxf_x = (y e^{xy}) \sin x + e^{xy} \cos x

  1. Differentiate (yexy)sinx(y e^{xy}) \sin x with respect to yy: y((yexy)sinx)=(exy+yxexy)sinx=exysinx+yxexysinx\frac{\partial}{\partial y} ((y e^{xy}) \sin x) = (e^{xy} + y x e^{xy}) \sin x = e^{xy} \sin x + y x e^{xy} \sin x
  2. Differentiate exycosxe^{xy} \cos x with respect to yy: y(exycosx)=xexycosx\frac{\partial}{\partial y} (e^{xy} \cos x) = x e^{xy} \cos x Combining these results, we get: fxy=exysinx+yxexysinx+xexycosxf_{xy} = e^{xy} \sin x + y x e^{xy} \sin x + x e^{xy} \cos x

Summary of Results

Thus, the mixed partial derivative fxyf_{xy} is: fxy=exysinx+yxexysinx+xexycosxf_{xy} = e^{xy} \sin x + y x e^{xy} \sin x + x e^{xy} \cos x

Would you like to see details for the other second partial derivatives or have any questions?

Here are 5 related questions to further your understanding:

  1. How would you find fyyf_{yy} for the same function?
  2. Why is it necessary to use the product rule in this case?
  3. What conditions guarantee that fxy=fyxf_{xy} = f_{yx}?
  4. How does the presence of exye^{xy} affect the derivatives?
  5. What would happen to the derivatives if we replaced sinx\sin x with cosx\cos x?

Tip: Mixed partial derivatives are often calculated by differentiating in a sequence. Just make sure each partial derivative is accurate before moving to the next.

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Math Problem Analysis

Mathematical Concepts

Calculus
Partial Derivatives
Product Rule

Formulas

Partial derivative of f(x, y) with respect to x or y
Product rule: (fg)' = f'g + fg'

Theorems

Clairaut's theorem on equality of mixed partial derivatives

Suitable Grade Level

College Calculus