1. implicit differentiation

Suppose that y is a function of x determined by e^xy = y. Find y(0) and y'(0).

2. Originally Posted by myoplex11
Suppose that y is a function of x determined by e^xy = y. Find y(0) and y'(0).
I'm assuming this is $e^{xy} = y$ and not $e^xy = y$ (which would be kind of silly, I guess)?

y(0):

$e^{0 \cdot y} = y$

$y = 1$

So y(0) = 1.

y'(0):
$e^{xy} = y$

$(y + xy')e^{xy} = y'$

So at x = 0 this becomes:
$(y(0) + 0 \cdot y')e^{0 \cdot y} = y'(0)$

$y(0) \cdot 1 = y'(0)$

Since y(0) = 1, thus y'(0) = 1 also.

-Dan