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July 30, 2015

July 30, 2015

Posted by **Anamika** on Tuesday, December 11, 2012 at 6:59am.

- differentiation -
**Reiny**, Tuesday, December 11, 2012 at 7:52amlooks like triple product rule

recall

d(uvw) = uvw' + uwv' + vwu'

I will also assume that you meant ln(x+y)

xy ln(x+y) = 1

xy(1/(x+y)) * (1 + dy/dx) + x ln(x+y) dy/dx + y ln(x+y) = 0

expand the first part to get at the dy/dx

xy(1/(x+y)) + dy/dx xy(1/(x+y)) + dy/dx (x ln(x+y)) = -y ln(x+y) - xy(1/(x+y))

dy/dx [xy(1/(x+y)) + (x ln(x+y)) ] = -y ln(x+y) - xy(1/(x+y))

dy/dx = (-y ln(x+y) - xy(1/(x+y))) / [xy(1/(x+y)) + (x ln(x+y)) ]

leave it like that or simplify as needed.