Find the equation of the tangent line at the indicated point.
y=5th root of (3x^3+4x) at (2,2)
y = (3x^3+4x)^(1/5)
y' = 1/5 (3x^3+4x)^(-4/5) * (9x^2+4)
y'(2) = (1/5)(1/16)(40) = 1/2
So, now you have a point and a slope. Go for it.
y=5th root of (3x^3+4x) at (2,2)
y' = 1/5 (3x^3+4x)^(-4/5) * (9x^2+4)
y'(2) = (1/5)(1/16)(40) = 1/2
So, now you have a point and a slope. Go for it.