# Thread: More implicit differentiation :(

1. ## More implicit differentiation :(

ln(xy) = 2 - x - y

The answer to this question in my book is dy/dx = [(-x - 1)/x][y/(1 + y)]

Could someone show me how to approach this question? I know I have to differentiate the ln(xy) term and apply the product rule to the (xy) inside, but I don't know how to do that exactly.

Thank!

2. Originally Posted by samstark
ln(xy) = 2 - x - y

The answer to this question in my book is dy/dx = [(-x - 1)/x][y/(1 + y)]

Could someone show me how to approach this question? I know I have to differentiate the ln(xy) term and apply the product rule to the (xy) inside, but I don't know how to do that exactly.

Thank!
Remember that the chain rule is also involved here. So you need to consider:
$\frac{d}{dx} f(g(x)) = \frac{df}{dg} \cdot \frac{dg}{dx}$

In this case f(u) = ln(u) and u = xy.
$\frac{d}{du} ln(u) = \frac{1}{u}$

$d(xy) = y + x \frac{dy}{dx}$

See where this gets you.

-Dan

3. You could also use the log laws on the left hand side

$\ln(ab) = \ln(a) + \ln(b)$