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If e^(xy) = 2, then what is dy/dx at the point (1, ln2)?
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If e^(xy) = 2, then what is dy/dx at the point (1, ln2)?

User Richmond Watkins
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e^(xy) = 2
(xdy/dx + y)e^(xy) = 0

At point (1, ln2), dy/dx + ln2 = 0
dy/dx = -ln2
User Lukus
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