Final answer:
The equation of the tangent line at the point (2,81), using the given derivative y' = 4(4x²-8x+3)(8x-8), is found to be y = 96x - 111 after substituting x=2 into the derivative to find the slope, which is 96, and applying the point-slope formula.
Step-by-step explanation:
The student asked to find the equation of the tangent line at the point (2,81) using the derivative y'=4(4x²-8x+3)(8x-8). To find the slope of the tangent line at that point, we substitute x=2 into the derivative.
Let's compute this:
y'(2) = 4(4(2)² - 8(2) + 3)(8(2) - 8) = 4(16 - 16 + 3)(16 - 8) = 4(3)(8) = 96.
The slope of the tangent line at point (2,81) is 96. Now, we use the point-slope form of a line, which is y - y1 = m(x - x1), where m is the slope and (x1, y1) is the point on the line.
Thus, the equation of the tangent line is:
y - 81 = 96(x - 2)
y = 96x - 192 + 81
y = 96x - 111
This is the equation of the tangent line at the point (2,81).