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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
\[(x^3-6x-1)'=3x^2-6\]
?

What did you get for the slope? I got a pretty little integer.

sorry. The answer is y=-9x+35

myininaya how did you get -9 and 3?

Your question said to find the tangent line at x=3.
I plugged x=3 into f'.

I did that and still did something wrong. Anyways, thanks for your help.

Hah, I found my mistake! I kept calculating 3^2 instead of 3^3 in the denominator...