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!

Quote:

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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!

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Remember that the chain rule is also involved here. So you need to consider:


In this case f(u) = ln(u) and u = xy.




See where this gets you.

-Dan

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