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I have one more similar ACT question. I asked you one before, about the Chain Rule; this is similar, but I'm not sure on how to apply that rule to finding the tangent line. I follow the process in the workbook, but keep getting different answers. Hoping you could help. Thanks.

It is:
Write an equation of the tangent line to the curve of y=(4x^(2)-2x)^3 at x=1

Hi John,
It would be more helpful if you showed me your progress. Anyway, following the same manner as before;
y = (4x - 2x)
Let u = 4x - 2x
y = u and dy/du = 3u
du/dx = 8x - 2

dy/dx = (dy/du).(du/dx)
= 3u . (8x - 2)
= 3(4x - 2x) . (8x - 2)

and at x = 1,
y = 8
dy/dx = 72   (which would be the slope of the tangent line at that point (1,8))

The equation is then, using the cartesian co-ordinates formula;
(y-8)/(x-1) = 72
y - 8 = 72(x - 1)
y - 8 = 72x - 72
y = 72x - 64


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Ahmed Salami


I can provide good answers to questions dealing in almost all of mathematics especially from A`Level downwards. I can as well help a good deal in Physics with most emphasis directed towards mechanics.


Aspiring theoretical physicist. I have been doing maths and physics all my life.

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