Determine an equation of the line tangent to the graph of
y= ((x^2-1)^2)/(x^3-6x-1) at the point (3,8)

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\[y=\frac{(x^2-1)^2}{x^3-6x-1}\]
Have you found y' yet?

So you have \[((x^2-1)^2)'=2(x^2-1) \cdot 2x \]
and
\[(x^3-6x-1)'=3x^2-6\]
?

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