Implicit Differentiation w/ Natural Log

**r2d2** · November 22nd 2009, 08:09 AM

$y + lnxy = 1$

Find $dy/dx$

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

But I do not know How to get there.

Thanks.

**skeeter** · November 22nd 2009, 08:22 AM

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 $\frac{dy}{dx}$

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