Urgent Help needed on finding tangent to a curve

[seostattler]

Find the the equation of a tangent to the curve where the tangent is at x=1 and the curve is; 4x^3 + 2/x^3 + lnx, i have not come across an equation of this nature before when working out tangents to curves and i am not sure of the method with i could use. i would be very grateful if any assitance could be provided with regards to this question.

[Glaysher]

y = 4x³ + 2x^-3 + ln x

dy/dx = 12x² - 6x^-4 + 1/x

Find gradient at x = 1

dy/dx = 12 - 6 + 1 = 7

Find y at x = 1

y = 4 + 2 + ln 1 = 6

So have m = 7, x₁ = 1, y₁ = 6

Stick into y - y₁ = m(x - x₁)

[seostattler]

thank you yes that is a lot simpler than i made it a rather daft error on my part thank you very much for your assistance