Stuck on Log Differentiation!! any help pleeease??

Question

Find an equation of the tangent line to the curve Y=In(X^2-2X-2) when X=3.

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Answer 1

y'= 1/(X^2-2X-2) (2x-2) = (2x-2)/ (X^2-2X-2)
so if x=3, y= ln( 9-6-2) =ln(1) =0
y'= (2(3) -2) / (4-4-2) = 4/-2= -2

the eq. of the tangent line is: y= -2 (x-3) .
or y= -2x + 6 .

Answer 2

y = ln(x^2 - 2x - 2)

Remember that the derivative of ln(x) is 1/x. That's what we do for here, and then we apply the chain rule.

y' = (1/[x^2 - 2x - 2]) (2x - 2), or
y' = [2x - 2]/[x^2 -2x - 2]

Therefore, y' is representative of slope, and we plug in x = 3 to solve for m.

m = [2(3) - 2]/[2^2 - 2(2) - 2] = [6 - 2]/[4 - 4 - 2] = 4/[-2] = -2

When x = 3, y = ln(3^2 - 2(3) - 2) = ln (9 - 6 - 2) = ln(1) = 0.

Therefore, we want to find the equation of a line with slope
m = -2 and through (3,0). This just became a grade 10 question.

We use the "slope = slope" method of obtaining the equation of the line; that is, we use the slope formula, with (x1,y1) = (3,0), and (x2,y2) = (x,y).

m = (y2 - y1)/(x2-x1)

But we know m = 2, and the rest of the values, so

2 = (y - 0)/(x - 3)
2 = y/(x - 3)
2(x - 3) = y
Therefore, y = 2x - 6.

I got a different answer from you ...