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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.

eeee
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  • Welcome to MSE! It really helps readability to write problems using MathJax (see FAQ). Regards – Amzoti Apr 19 '13 at 15:40
  • If someone can edit this for me,until I learn how to edit properly,I will be very grateful :) – eeee Apr 19 '13 at 15:40
  • FYI: when you receive answers that are helpful, you may accept one answer per question. You can accept an answer by clicking on the "$\large \checkmark$" symbol located to the left of the answer you'd like to accept. Plus, you get two reputation points for every answer you accept. – amWhy Apr 20 '13 at 22:45

2 Answers2

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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.

Dennis Gulko
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  • No,I need this with the first one,but when I replace what I got,I dont get it right :/ – eeee Apr 19 '13 at 16:34
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You could just use the chain rule as follows:

$xyz=f(t)=\cos(t)t^{-1}e^t$

$(\frac{d}{dt}U(f(t)))=(\frac{d}{df}U(f))(\frac{d}{dt}f(t))$

${\frac {d}{df}}U \left( f \right) ={\frac {d}{df}}\arctan \left( f \right)= \left( 1+{f}^{2} \right) ^{-1}$

${\frac {d}{dt}}f \left( t \right) =(-{\frac {\sin \left( t \right) }{t} }+{\frac {\cos \left( t \right) }{t}}-{\frac {\cos \left( t \right) }{ {t}^{2}}} )e^{t}$ ...use product rule here.

${\frac {d}{dt}}U \left( t \right) =-{\frac {{{\rm e}^{t}} \left( \sin \left( t \right) t-\cos \left( t \right) t+\cos \left( t \right) \right) }{{t}^{2}+ \left( \cos \left( t \right) \right) ^{2}{{\rm e} ^{2\,t}}}} $

Graham Hesketh
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