# Quotient & Product Rule: Same funtion, different answers?

1. Feb 22, 2014

### AntSC

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: Feb 22, 2014
2. Feb 22, 2014

### D H

Staff Emeritus
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. Feb 22, 2014

### AntSC

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?

4. Feb 22, 2014

### D H

Staff Emeritus
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. Feb 22, 2014

### AntSC

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