differentiating ln and efunctions

**200001** · August 10th 2010, 12:10 AM

How would I find out dy/dy if:

y = e^xy

and another different question

y = ln(xy)

Thanks

**Prove It** · August 10th 2010, 12:15 AM

Are you trying to find $\frac{dy}{dx}$ ?

1. $y = e^xy$

$\frac{dy}{dx} = \frac{d}{dx}(e^{xy})$

$\frac{dy}{dx} = e^{xy}\frac{d}{dx}(xy)$

$\frac{dy}{dx} = e^{xy}\left(x\frac{dy}{dx} + y\right)$

$\frac{dy}{dx} = xe^{xy}\frac{dy}{dx} + ye^{xy}$

$\frac{dy}{dx} - xe^{xy}\frac{dy}{dx} = ye^{xy}$

$\frac{dy}{dx}(1 - xe^{xy}) = ye^{xy}$

$\frac{dy}{dx} = \frac{ye^{xy}}{1 - xe^{xy}}$ .

**Prove It** · August 10th 2010, 12:17 AM

Originally Posted by 200001

How would I find out dy/dy if:

y = e^xy

and another different question

y = ln(xy)

Thanks

2. $y = \ln{(xy)}$

$e^y = xy$

$\frac{d}{dx}(e^y) = \frac{d}{dx}(xy)$

$e^y\frac{dy}{dx} = x\frac{dy}{dx} + y$

$e^y\frac{dy}{dx} - x\frac{dy}{dx} = y$

$\frac{dy}{dx}(e^y - x) = y$

$\frac{dy}{dx} = \frac{y}{e^y - x}$ .

**HallsofIvy** · August 10th 2010, 02:33 AM

Different way to do the same thing:
$y= ln(xy)$
$\frac{dy}{dx}= \frac{1}{xy}\frac{d(xy)}{dx}$
$\frac{dy}{dx}= \frac{1}{xy}\left(y+ x\frac{dy}{dx}\right)$
$\frac{dy}{dx}= \frac{1}{xy}(y)+ \frac{1}{xy}\left(x\frac{dy}{dx}\right)$
$\frac{dy}{dx}= \frac{1}{x}+ \frac{1}{y}\frac{dy}{dx}$ \
$\frac{dy}{dx}- \frac{1}{y}\frac{dy}{dx}= \frac{1}{x}$
$\frac{dy}{dx}\left(1- \frac{1}{y}\right)= \frac{1}{x}$
$\frac{dy}{dx}= \frac{\frac{1}{x}}{1- \frac{1}{y}}= \frac{y}{xy- x}$

This may look different from Prove It's result but remember that y= ln(xy) so that $xy= e^y$ . The "xy" in my denominator is the same as the " $e^y$ " in his.

**Vlasev** · August 10th 2010, 04:13 AM

2nd method for the first one:

$\displaystyle y = e^{xy}$

$\displaystyle \ln(y) = xy$

$\displaystyle \frac{1}{y}\frac{dy}{dx} = y + x \frac{dy}{dx}$

$\displaystyle \frac{1-xy}{y}\frac{dy}{dx} = y$

Hence

$\displaystyle \frac{dy}{dx} = \frac{y^2}{1-xy}$

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.

**200001** · August 11th 2010, 08:26 PM

Thanks to all those who have posted. Between them I have grasped it !

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