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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?
Yes, I have but I am getting the final answer wrong, and I am not sure where I am making a mistake. It is the first time I can't figure out what I am doing wrong...
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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So by quotient rule, we should have \[y'=\frac{2(x^2-1) \cdot 2x (x^3-6x-1)-(3x^2-6)(x^2-1)^2}{(x^3-6x-1)^2}\]
What did you get for the slope? I got a pretty little integer.
Say something so I can know you are still there. Like did you simplify that above and then plug in? I'm getting the slope of the tangent at x=3 is -9.
sorry. The answer is y=-9x+35
I got some bizzare answers that don't make much sense... However, for y' i got \[((x^2-1)(x^4-15x^2-4x-6))\div(x^3-6x-1)^2\]
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.
You y' is right... I don't understand why you aren't getting -9 when pluggin in 3.... \[\frac{(3^2-1)(3^4-15(3)^2-4(3)-6)}{(3^3-6(3)-1)^2}\] \[=\frac{(9-1)(81-135-12-6)}{(27-18-1)^2}=\frac{(8)(81-153)}{(27-19)^2}\] \[=\frac{(8)(-72)}{(8)^2}=\frac{8(-72)}{(8)(8)}=\frac{\cancel{8}(-72)}{\cancel{(8)}(8)}=\frac{-72}{8}=-9\]
You should be getting -9 as the slope of your line... So I don't know where you are going wrong in your calculation.
Hah, I found my mistake! I kept calculating 3^2 instead of 3^3 in the denominator...

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