Determine the equation of the tangent to h(x)=2lnx/3xat the point where x=1
h(x) = 2lnx / (3x)
h ' (x) = [ (3x)(2/x) - 3(2lnx) ]/(9x^2) , using the quotient rule
= (6 - 6lnx)/(9x^2)
when x = 1,
h(1) = 2ln 1/3 = 0
so our point is (1,0)
h ' (1) = (6 - 0)/9 = 2/3
y - 0 = (2/3)(x - 1)
change to whatever form you need