$\displaystyle y + lnxy = 1$ Find $\displaystyle dy/dx$ The answer is $\displaystyle -y/(x(y+1))$ But I do not know How to get there. Thanks.
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Originally Posted by r2d2 $\displaystyle y + lnxy = 1$ Find $\displaystyle dy/dx$ The answer is $\displaystyle -y/(x(y+1))$ But I do not know How to get there. Thanks. $\displaystyle y + \ln(xy) = 1$ $\displaystyle y + \ln(x) + \ln(y) = 1$ $\displaystyle \frac{dy}{dx} + \frac{1}{x} + \frac{1}{y} \cdot \frac{dy}{dx} = 0$ finish the algebra and solve for $\displaystyle \frac{dy}{dx}$
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