Math question partial derivatives help? I have the functions : $u=\arctan(xyz)$ where $x=\cos(t)\quad  y=e^t$ and $z=1/t$.
I have to find $\dfrac{du}{dt}$.
My attempt to a solution : $\dfrac{du}{dt}=\dfrac{\partial u}{\partial x}\dfrac{dx}{dt} + \dfrac{\partial u}{\partial y}\dfrac{dy}{dt} + \dfrac{\partial u}{\partial z}\dfrac{dz}{dt}$
$\dfrac{\partial u}{\partial x}=\dfrac{yz}{1+x^2\cdot y^2\cdot z^2}$ ;$\dfrac{dx}{dt}=\dfrac{-\sin t \partial u}{\partial y}=\dfrac{xz}{1+x^2\cdot y^2\cdot z^2}$ ,$\dfrac{dy}{dt}=e^t$ ,
$\dfrac{\partial u}{\partial z}=\dfrac{xy}{1+x^2\cdot y^2 \cdot z^2}$ $\dfrac{dz}{dt}=\dfrac{-1}{t^2}$.
The problem is,how do I replace these back into the formula so I could get a result : 
$\dfrac{e^t(t\cos t-t\sin t+\cos t)}{t^2+e^{(2t)} \cdot \cos^2(t)}$
I SOLVED THIS,THANK YOU ALREADY.

You did everything right. There are two ways to finish this:
1) Continue from what you got: replace $x,y,z$ by the functions of $t$ and get the common denumerator.
2) Write $U(t)=\arctan(xyz)=\arctan\left(\frac{e^t\cos t}t\right)$. Now just compute the derivative.
