Write an equation of the line tangent to y=(9-x^2)^2/3 at x=1

Answer 1

dy/dx = (2/3)(9-x^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