How would I find out dy/dy if:
y = e^xy
and another different question
y = ln(xy)
Thanks
Are you trying to find $\displaystyle \frac{dy}{dx}$?
1. $\displaystyle y = e^xy$
$\displaystyle \frac{dy}{dx} = \frac{d}{dx}(e^{xy})$
$\displaystyle \frac{dy}{dx} = e^{xy}\frac{d}{dx}(xy)$
$\displaystyle \frac{dy}{dx} = e^{xy}\left(x\frac{dy}{dx} + y\right)$
$\displaystyle \frac{dy}{dx} = xe^{xy}\frac{dy}{dx} + ye^{xy}$
$\displaystyle \frac{dy}{dx} - xe^{xy}\frac{dy}{dx} = ye^{xy}$
$\displaystyle \frac{dy}{dx}(1 - xe^{xy}) = ye^{xy}$
$\displaystyle \frac{dy}{dx} = \frac{ye^{xy}}{1 - xe^{xy}}$.
2. $\displaystyle y = \ln{(xy)}$
$\displaystyle e^y = xy$
$\displaystyle \frac{d}{dx}(e^y) = \frac{d}{dx}(xy)$
$\displaystyle e^y\frac{dy}{dx} = x\frac{dy}{dx} + y$
$\displaystyle e^y\frac{dy}{dx} - x\frac{dy}{dx} = y$
$\displaystyle \frac{dy}{dx}(e^y - x) = y$
$\displaystyle \frac{dy}{dx} = \frac{y}{e^y - x}$.
Different way to do the same thing:
$\displaystyle y= ln(xy)$
$\displaystyle \frac{dy}{dx}= \frac{1}{xy}\frac{d(xy)}{dx}$
$\displaystyle \frac{dy}{dx}= \frac{1}{xy}\left(y+ x\frac{dy}{dx}\right)$
$\displaystyle \frac{dy}{dx}= \frac{1}{xy}(y)+ \frac{1}{xy}\left(x\frac{dy}{dx}\right)$
$\displaystyle \frac{dy}{dx}= \frac{1}{x}+ \frac{1}{y}\frac{dy}{dx}$\
$\displaystyle \frac{dy}{dx}- \frac{1}{y}\frac{dy}{dx}= \frac{1}{x}$
$\displaystyle \frac{dy}{dx}\left(1- \frac{1}{y}\right)= \frac{1}{x}$
$\displaystyle \frac{dy}{dx}= \frac{\frac{1}{x}}{1- \frac{1}{y}}= \frac{y}{xy- x}$
This may look different from Prove It's result but remember that y= ln(xy) so that $\displaystyle xy= e^y$. The "xy" in my denominator is the same as the "$\displaystyle e^y$" in his.
2nd method for the first one:
$\displaystyle \displaystyle y = e^{xy}$
$\displaystyle \displaystyle \ln(y) = xy$
$\displaystyle \displaystyle \frac{1}{y}\frac{dy}{dx} = y + x \frac{dy}{dx}$
$\displaystyle \displaystyle \frac{1-xy}{y}\frac{dy}{dx} = y$
Hence
$\displaystyle \displaystyle \frac{dy}{dx} = \frac{y^2}{1-xy}$
This method is known as logarithmic differentiation is MUCH easier to use in if you have complicated product on the right hand side.
P.S. this problem should have been posted in the Calculus section as it involves derivatives.