Quotient & Product Rule: Same funtion, different answers?

In summary, 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
  • #1
AntSC
65
3
Can someone check my working. I don't understand why i am getting different answers?
[tex]u(x,t)=\frac{{e}^{-\frac{x^2}{4Dt}}}{\sqrt{4Dt}}[/tex]
Differentiate w.r.t 't' by quotient rule:
[tex]\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}[/tex]
[tex]\frac{\partial u}{\partial t}=\left[ \frac{x^2}{4Dt^2}u(x,t)+\frac{1}{2t}u(x,t)\right]\frac{1}{4Dt}[/tex]
For the product rule, write as -
[tex]u(x,t)={\left(4D \right)}^{-\frac{1}{2}}{t}^{-\frac{1}{2}}{e}^{-\frac{x^2}{4Dt}}[/tex]
Differentiate w.r.t 't' by product rule:
[tex]\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][/tex]
[tex]\frac{\partial u}{\partial t}=\frac{x^2}{4Dt^2}u(x,t)-\frac{1}{2t}u(x,t)[/tex]
This is clearly not the same.
What am i doing wrong?
 
Last edited:
Physics news on Phys.org
  • #2
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
$$\frac{\partial u(x,t)}{\partial t} = \frac{\frac{\partial v(x,t)}{\partial t}w(t) - v(x,t)\frac{\partial w(t)}{\partial t}}{w(t)^2}$$
 
  • #3
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 [itex]v(x,t)[/itex] and [itex]w(t)[/itex] is what i chose for the quotient rule. The [itex]w(t)^2[/itex] is the [itex]\frac {1}{4Dt}[/itex] term outside the brackets.
I don't see what am i missing?
 
  • #4
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.
 
  • #5
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
 

1. What is the quotient rule?

The quotient rule is a mathematical formula used to find the derivative of a function that is the ratio of two other functions. It states that the derivative of f(x) divided by g(x) is equal to the derivative of f(x) multiplied by g(x) minus the derivative of g(x) multiplied by f(x), all divided by the square of g(x).

2. How is the quotient rule different from the product rule?

The quotient rule is used to find the derivative of a function that is a ratio, while the product rule is used to find the derivative of a function that is a product of two other functions. The formula for the quotient rule involves subtraction and division, while the formula for the product rule only involves multiplication.

3. What does it mean when the quotient and product rule give different answers?

This means that the function in question is not a simple product or ratio of two other functions. In other words, it is a composite function or a function that is a combination of multiple other functions. In such cases, the quotient and product rule cannot be applied directly and more advanced techniques may be needed to find the derivative.

4. Can the quotient and product rule be used interchangeably?

No, the quotient and product rule cannot be used interchangeably. They are specific rules that are applied to different types of functions. Using the wrong rule will result in an incorrect answer. It is important to understand the difference between the two and when to use each one.

5. Are there any real-life applications of the quotient and product rule?

Yes, the quotient and product rule are used in various fields of science and engineering to find rates of change and optimize functions. For example, in physics, the quotient rule can be used to find the acceleration of an object given its position and velocity functions. In economics, the product rule can be used to find the optimal production level for a company given its cost and revenue functions.

Similar threads

  • Calculus and Beyond Homework Help
Replies
4
Views
694
  • Calculus and Beyond Homework Help
Replies
5
Views
620
Replies
4
Views
648
  • Calculus and Beyond Homework Help
Replies
1
Views
160
  • Calculus and Beyond Homework Help
Replies
3
Views
495
  • Calculus and Beyond Homework Help
Replies
8
Views
237
  • Calculus and Beyond Homework Help
Replies
6
Views
761
  • Calculus and Beyond Homework Help
Replies
22
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
741
  • Calculus and Beyond Homework Help
Replies
12
Views
993
Back
Top