Math Problem Statement

z = x sin(xy) ∂z ∂x

Incorrect: Your answer is incorrect. ∂z ∂y

Solution

To find the partial derivatives of the given function:

z=xsin(xy)z = x \sin(xy)

we'll differentiate zz with respect to xx and yy.

1. Partial derivative of zz with respect to xx:

We treat yy as a constant while differentiating with respect to xx.

zx=x(xsin(xy))\frac{\partial z}{\partial x} = \frac{\partial}{\partial x} \left( x \sin(xy) \right)

Using the product rule:

zx=sin(xy)+xx(sin(xy))\frac{\partial z}{\partial x} = \sin(xy) + x \cdot \frac{\partial}{\partial x} \left( \sin(xy) \right)

Now, applying the chain rule to differentiate sin(xy)\sin(xy) with respect to xx:

x(sin(xy))=cos(xy)y\frac{\partial}{\partial x} \left( \sin(xy) \right) = \cos(xy) \cdot y

Therefore, the partial derivative with respect to xx is:

zx=sin(xy)+xcos(xy)y\frac{\partial z}{\partial x} = \sin(xy) + x \cos(xy) \cdot y

2. Partial derivative of zz with respect to yy:

Now, treat xx as a constant while differentiating with respect to yy:

zy=y(xsin(xy))\frac{\partial z}{\partial y} = \frac{\partial}{\partial y} \left( x \sin(xy) \right)

Using the chain rule:

zy=xcos(xy)x\frac{\partial z}{\partial y} = x \cdot \cos(xy) \cdot x

This simplifies to:

zy=x2cos(xy)\frac{\partial z}{\partial y} = x^2 \cos(xy)

Summary of Results:

  • zx=sin(xy)+xycos(xy)\frac{\partial z}{\partial x} = \sin(xy) + x \cdot y \cos(xy)
  • zy=x2cos(xy)\frac{\partial z}{\partial y} = x^2 \cos(xy)

Let me know if you want more details or further explanations.


Here are some related questions for you to explore:

  1. What is the second partial derivative of zz with respect to xx?
  2. How would you calculate the mixed partial derivative 2zxy\frac{\partial^2 z}{\partial x \partial y}?
  3. What would happen if you took the derivative of zz with respect to time, assuming xx and yy are both functions of time?
  4. Can you generalize the method used for partial derivatives to higher-order functions of multiple variables?
  5. How do partial derivatives relate to the concept of tangent planes in multivariable calculus?

Tip: When using the chain rule in partial derivatives, always carefully distinguish between the variable you're differentiating with respect to and the others that are treated as constants.

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Partial Derivatives
Multivariable Calculus
Product Rule
Chain Rule

Formulas

∂z/∂x = sin(xy) + x * y * cos(xy)
∂z/∂y = x^2 * cos(xy)

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Grades 11-12