We have y - ln ( 1/y) + ln y = ln
x
use the relation that log a + log b = log
a*b
=> y - ln [( 1/y)(1/y)] = ln
x
=> y - ln [ 1/y^2] = ln
x
=> dy/dx - y^2*(-2)*(1/y^3)*dy/dx =
1/x
=> dy/dx [ 1 + 2* y^2 / y^3] =
1/x
=> dy/dx [ 1 + 2/y]=
1/x
=> dy/dx = 1/[x*(1 +
2/y)]
The required value of dy/dx = 1/[x*(1
+ 2/y)]
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