How do you implicitly differentiate $ln(xy)=x+y$ ?

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Answer 1 · 2016-07-01T11:43:23

$(xy-y)/(x-xy)$

Given expression

$ln(xy)=x+y$

$=>lnx+lny=x+y$

Differentiating w.r. to x we can write

$(d(lnx))/(dx)+(d(lny))/(dx)=(d(x))/(dx)+(d(y))/(dx)$

$=>1/x+1/y*(dy)/(dx)=1+(dy)/(dx)$

$=>1/y*(dy)/(dx)-(dy)/(dx)=1-1/x$

$=>(1/y-1)(dy)/(dx)=(x-1)/x$

$=>((1-y)/y)(dy)/(dx)=(x-1)/x$

$=>(dy)/(dx)=(x-1)/x xx(y)/(1-y)=(xy-y)/(x-xy)$

How do you implicitly differentiate ln(xy)=x+y ln(xy)=x+y ?