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