What is the partial derivative of f(x, y) with respect to x?

  • Context: MHB 
  • Thread starter Thread starter Kris1
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary
SUMMARY

The discussion focuses on calculating the partial derivatives of the function f(x, y) = 2y³ + xy² + (7/2)y² + x². The correct partial derivative with respect to x is ∂f/∂x = 2xy + 2x, and with respect to y is ∂f/∂y = 6y² + x² + 2y. Participants emphasized the importance of treating the other variable as a constant when differentiating. The final corrections were confirmed by multiple users, ensuring clarity in the differentiation process.

PREREQUISITES
  • Understanding of partial derivatives in multivariable calculus
  • Familiarity with the notation and rules of differentiation
  • Basic knowledge of polynomial functions
  • Experience with mathematical expressions and simplification
NEXT STEPS
  • Study the concept of gradient vectors in multivariable calculus
  • Learn about the chain rule for partial derivatives
  • Explore applications of partial derivatives in optimization problems
  • Investigate the use of software tools like Wolfram Alpha for symbolic differentiation
USEFUL FOR

Students of calculus, mathematicians, and anyone interested in understanding multivariable functions and their derivatives.

Kris1
Messages
29
Reaction score
0
Hi I am trying to find the gradient function of
f(x, y) = 2*y^3+y*x*2+7/2 y^2+x^2

I have worked out the answer to be
d/dx = 2*y^3+2*y*x+7/2*y^2+2*x
d/dy = 6*y^2+x^2+(7/2)*2*y+x^2

Can someone please check this to make sure I have done the partial derivatives correctly as I am unsure if they are right?

EDIT:

Excuse my poor working the correct answer when I resolved is
2*x*y+2*x,6*y^2+x^2+2*y
 
Last edited:
Physics news on Phys.org
Those are not correct...you need to treat the terms involving the other variable as a constant when taking the partial with respect to a variable/ For example, if given:

$$f(x,y)=x^2+xy+y^2$$

then:

$$\frac{\partial f}{\partial x}=2x+y$$

$$\frac{\partial f}{\partial y}=x+2y$$

Do you see how the terms that do not involve the variable with respect to which we are differentiating vanish to zero?
 
Hi Mark,

I realized my error and corrected it as your post went up. Thanks for the reminder though
 
Kris said:
Hi Mark,

I realized my error and corrected it as your post went up. Thanks for the reminder though

Your results are still not quite right. To make sure, I think you mean to state:

$$f(x,y)=2y^3+x^2y+\frac{7}{2}y^2+x^2$$

Is this correct?
 
Yes that is what I meant
 
Okay, good! :D

Now, let's find the partial with respect to $x$ first, treating $y$ as a constant...what do you get?
 

Similar threads

  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 8 ·
Replies
8
Views
898
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 12 ·
Replies
12
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 3 ·
Replies
3
Views
3K
  • · Replies 16 ·
Replies
16
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 11 ·
Replies
11
Views
2K