# Thread: Implicit Differentiation w/ Natural Log

1. ## Implicit Differentiation w/ Natural Log

$y + lnxy = 1$

Find $dy/dx$

The answer is $-y/(x(y+1))$

But I do not know How to get there.

Thanks.

2. Originally Posted by r2d2
$y + lnxy = 1$

Find $dy/dx$

The answer is $-y/(x(y+1))$

But I do not know How to get there.

Thanks.
$y + \ln(xy) = 1$

$y + \ln(x) + \ln(y) = 1$

$\frac{dy}{dx} + \frac{1}{x} + \frac{1}{y} \cdot \frac{dy}{dx} = 0$

finish the algebra and solve for $\frac{dy}{dx}$