Math Problem Statement

For the function

​f(x,y)equals=x squared e Superscript 3 xyx3e2xy​,

find

f Subscript x

and

f Subscript y.

Solution

To compute the partial derivatives fxf_x and fyf_y of the function

f(x,y)=x2e3xy,f(x, y) = x^2 e^{3xy},

we will differentiate f(x,y)f(x, y) with respect to xx and yy, treating the other variable as a constant.


1. Partial derivative with respect to xx (fxf_x):

The product rule is required because f(x,y)f(x, y) is a product of x2x^2 and e3xye^{3xy}. The product rule states:

x(uv)=uv+uv,\frac{\partial}{\partial x}(u v) = u'v + uv',

where u=x2u = x^2 and v=e3xyv = e^{3xy}.

Step 1: Differentiate u=x2u = x^2 with respect to xx:

x(x2)=2x.\frac{\partial}{\partial x}(x^2) = 2x.

Step 2: Differentiate v=e3xyv = e^{3xy} with respect to xx:

Using the chain rule: x(e3xy)=e3xyx(3xy)=e3xy3y.\frac{\partial}{\partial x}(e^{3xy}) = e^{3xy} \cdot \frac{\partial}{\partial x}(3xy) = e^{3xy} \cdot 3y.

Step 3: Apply the product rule:

fx=x(x2e3xy)=(2x)e3xy+(x2)(3ye3xy).f_x = \frac{\partial}{\partial x}(x^2 e^{3xy}) = (2x)e^{3xy} + (x^2)(3y e^{3xy}).

Simplify: fx=2xe3xy+3x2ye3xy.f_x = 2x e^{3xy} + 3x^2 y e^{3xy}.


2. Partial derivative with respect to yy (fyf_y):

We use the product rule again, where u=x2u = x^2 and v=e3xyv = e^{3xy}.

Step 1: Differentiate u=x2u = x^2 with respect to yy:

Since x2x^2 does not depend on yy: y(x2)=0.\frac{\partial}{\partial y}(x^2) = 0.

Step 2: Differentiate v=e3xyv = e^{3xy} with respect to yy:

Using the chain rule: y(e3xy)=e3xyy(3xy)=e3xy3x.\frac{\partial}{\partial y}(e^{3xy}) = e^{3xy} \cdot \frac{\partial}{\partial y}(3xy) = e^{3xy} \cdot 3x.

Step 3: Apply the product rule:

fy=y(x2e3xy)=(0)e3xy+(x2)(3xe3xy).f_y = \frac{\partial}{\partial y}(x^2 e^{3xy}) = (0)e^{3xy} + (x^2)(3x e^{3xy}).

Simplify: fy=3x3e3xy.f_y = 3x^3 e^{3xy}.


Final Results:

fx=2xe3xy+3x2ye3xy,f_x = 2x e^{3xy} + 3x^2 y e^{3xy}, fy=3x3e3xy.f_y = 3x^3 e^{3xy}.

Let me know if you’d like a step-by-step explanation of any part of this!


Questions to Expand:

  1. How does the chain rule apply when differentiating exponentials?
  2. What is the significance of partial derivatives in multivariable functions?
  3. Can we apply these results to evaluate fxf_x and fyf_y at specific points (e.g., x=1,y=0x=1, y=0)?
  4. How do we visualize partial derivatives geometrically?
  5. What are higher-order partial derivatives for f(x,y)f(x, y)?

Tip:

Always keep track of which variable is being treated as constant when taking partial derivatives!

Ask a new question for Free

By Image

Drop file here or Click Here to upload

Math Problem Analysis

Mathematical Concepts

Multivariable Calculus
Partial Derivatives
Chain Rule
Product Rule

Formulas

Product Rule: ∂(uv)/∂x = u'v + uv'
Chain Rule: ∂(e^g)/∂x = e^g * ∂g/∂x

Theorems

Chain Rule
Product Rule

Suitable Grade Level

College-level Calculus