calculus
posted by Amie .
Write an equation of the line tangent to y=(9x^2)^2/3 at x=1

dy/dx = (2/3)(9x^2)^(1/3) (2x)
when x = 1, y = 8^(2/3) = 4
dy/dx = (2/3)(8)^(1/3) (2)
= (4/3)((1/2) = 2/3
equation:
y = (2/3)x + b
at( 1,4)
4 = (2/3) + b
b = 10/3
y = (2/3)x + 10/3