MHB Calc Solved! Get Quick Help With Your Calculus Q

  • Thread starter Thread starter AHMED2021
  • Start date Start date
  • Tags Tags
    Calculus
Click For Summary
The discussion revolves around solving a calculus problem involving partial derivatives using the chain rule. It confirms that f is a function of x, which in turn is a function of u and v. The derivatives are expressed as $\frac{\partial f}{\partial u}= \frac{df}{dx}\frac{\partial x}{\partial u}$ and $\frac{\partial f}{\partial v}= \frac{df}{dx}\frac{\partial x}{\partial v}$. The function f(x) is defined as arctan(x), with its derivative $\frac{df}{dx}= \frac{1}{x^2+ 1}$. The variable x is expressed as $x= e^u+ ln(v)$, leading to the partial derivatives $\frac{\partial x}{\partial u}= e^u$ and $\frac{\partial x}{\partial v}= \frac{1}{v}$.
AHMED2021
Messages
1
Reaction score
0
please I need solve of this question very quickly
 

Attachments

  • upload_2021_03_24_07_26_54_427.jpg
    upload_2021_03_24_07_26_54_427.jpg
    203.7 KB · Views: 111
Physics news on Phys.org
Then you had better get started! Can we at least assume you are taking a Calculus class?

Do you know the "chain rule"?

f is a function of x and x itself is a function of u and v so
$\frac{\partial f}{\partial u}= \frac{df}{dx}\frac{\partial x}{\partial u}$ and $\frac{\partial f}{\partial v}= \frac{df}{dx}\frac{\partial x}{\partial v}$.

f(x)= arctan(x) so $\frac{df}{dx}= \frac{1}{x^2+ 1}$

$x= e^u+ ln(v)$ so $\frac{\partial x}{\partial u}= e^u$ and $\frac{\partial x}{\partial v}= \frac{1}{v}$.
 
Last edited:

Similar threads

  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 10 ·
Replies
10
Views
2K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 3 ·
Replies
3
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 6 ·
Replies
6
Views
4K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
5
Views
1K