Quotient & Product Rule: Same funtion, different answers?

Click For Summary

Homework Help Overview

The discussion revolves around the differentiation of a function \( u(x,t) = \frac{e^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}} \) with respect to \( t \) using both the quotient and product rules. Participants are exploring the discrepancies in the results obtained from these two methods.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • The original poster attempts to differentiate the function using both the quotient and product rules, noting that the results differ. Some participants question the correctness of the application of the quotient rule and suggest re-evaluating the differentiation steps.

Discussion Status

Participants are actively engaging in identifying potential errors in the original poster's differentiation process. Some have pointed out specific mistakes related to the application of the quotient rule and the differentiation of the denominator. The discussion is ongoing, with hints of resolution as the original poster acknowledges a mistake.

Contextual Notes

There is a focus on ensuring the correct application of differentiation rules, with specific attention to the handling of radical terms and the components of the quotient rule. The original poster expresses frustration over the confusion caused by the differing results.

AntSC
Messages
65
Reaction score
3
Can someone check my working. I don't understand why i am getting different answers?
u(x,t)=\frac{{e}^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}}
Differentiate w.r.t 't' by quotient rule:
\frac{\partial u}{\partial t}=\left[ \frac{1}{\sqrt{4Dt}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}}-{e}^{-\frac{x^2}{4Dt}}\cdot \frac{1}{\sqrt{4D}}\cdot \frac{-1}{2t\sqrt{t}} \right]\frac{1}{4Dt}
\frac{\partial u}{\partial t}=\left[ \frac{x^2}{4Dt^2}u(x,t)+\frac{1}{2t}u(x,t)\right]\frac{1}{4Dt}
For the product rule, write as -
u(x,t)={\left(4D \right)}^{-\frac{1}{2}}{t}^{-\frac{1}{2}}{e}^{-\frac{x^2}{4Dt}}
Differentiate w.r.t 't' by product rule:
\frac{\partial u}{\partial t}=\frac{1}{\sqrt{4D}}\left[ {e}^{-\frac{x^2}{4Dt}}\cdot \frac{-1}{2t\sqrt{t}}+\frac{1}{\sqrt{t}}\cdot \frac{x^2}{4Dt^2}\cdot {e}^{-\frac{x^2}{4Dt}} \right]
\frac{\partial u}{\partial t}=\frac{x^2}{4Dt^2}u(x,t)-\frac{1}{2t}u(x,t)
This is clearly not the same.
What am i doing wrong?
 
Last edited:
Physics news on Phys.org
You didn't use the quotient rule correctly. Denoting ##u(x,t)=\frac{v(x,t)}{w(t)}## where ##v(x,t)=\exp\left(\frac {-x^2}{4Dt}\right)## and ##w(t)=\sqrt{4Dt}##, then the quotient rule says
$$\frac{\partial u(x,t)}{\partial t} = \frac{\frac{\partial v(x,t)}{\partial t}w(t) - v(x,t)\frac{\partial w(t)}{\partial t}}{w(t)^2}$$
 
D H said:
You didn't use the quotient rule correctly. Denoting ##u(x,t)=\frac{v(x,t)}{w(t)}## where ##v(x,t)=\exp\left(\frac {-x^2}{4Dt}\right)## and ##w(t)=\sqrt{4Dt}##, then the quotient rule says
$$\frac{\partial u(x,t)}{\partial t} = \frac{\frac{\partial v(x,t)}{\partial t}w(t) - v(x,t)\frac{\partial w(t)}{\partial t}}{w(t)^2}$$
The terms you wrote for v(x,t) and w(t) is what i chose for the quotient rule. The w(t)^2 is the \frac {1}{4Dt} term outside the brackets.
I don't see what am i missing?
 
You did a couple of things wrong. You computed ##\frac{d}{dt}\sqrt{4Dt}## incorrectly, and you lost a radical term somewhere along the line in your quotient rule computations.
 
D H said:
You did a couple of things wrong. You computed ##\frac{d}{dt}\sqrt{4Dt}## incorrectly, and you lost a radical term somewhere along the line in your quotient rule computations.

I've spotted my stupid mistake. Thanks for the pointer. I've been going round in circles with this.
Cheers
 

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 105 ·
4
Replies
105
Views
10K
Replies
5
Views
2K
Replies
4
Views
2K
Replies
4
Views
1K
  • · Replies 11 ·
Replies
11
Views
3K
Replies
6
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
3K