This forum is such a good resource - I have another question regarding implicit differentiation that I can't find any information in my notes for.

I have the following question...

Assume x and y are functions of t. Evaluate $\displaystyle \frac {dy}{dt}$ for the following.

$\displaystyle xe^{y} = 2 - ln(2) + ln(x) ....... [ \frac {dx}{dt}= 6 , x = 2 , y = 0]$

What I'm a little unsure about, is that when working through this, mainly when using the chain rule for exponential functions when trying to find $\displaystyle \frac {dy}{dt}$

I got up to about the second step shown below, but then realized that something isn't right at all. I completed the equation to it's (incorrect) conclusion, but I don't really no how to handle the exponents and Euler's number when working with it.

$\displaystyle \frac {d}{dt} (xe^y) = \frac {d}{dt}(2) - \frac {d}{dt} ln(2) + \frac {d}{dt} ln(x)$

$\displaystyle = [x (ye^y \frac{dy}{dt}) + e^y( \frac{dx}{dt}] = 0 - \frac {0}{2} + \frac {1 \frac {dx}{dt}}{x} $

$\displaystyle = xye^y \frac {dy}{dt} + e^y \frac{dx}{dt} = \frac {1 \frac {dx}{dt}}{x}$

$\displaystyle = xye^y \frac {dy}{dt} = \frac {1 \frac{dx}{dt}} {x} - e^y \frac{dx}{dt}$

=$\displaystyle \frac {dy}{dt} = \frac {1 \frac {dx}{dt}}{x^2ye^y} - \frac {ey\frac {dy}{dx}}{xye^y}$

I'm pretty sure that I've already made a mistake somewhere along the line here, but just to make sure...

$\displaystyle \frac {dy}{dt} = \frac {1(6)}{4)(0)(1)} - \frac {e(0)(6)}{(2)(0)(1)}$

$\displaystyle = \frac {6}{0} - 0$

Which is just ridiculous...