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Find an equation of the tangent line to the curve at the given point.y = ln(x2 − 6x + 1),    (6, 0)

Question

Find an equation of the tangent line to the curve at the given point.

y = ln(x2 − 6x + 1),    (6, 0)
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Step 1

Equation of the line to the curve at the given point is shown below.

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Step 2

Differentiate the given function with respect to x and the slope at the given point (6, 0) is calculated as follows.

фу
1
d x2-6x+1 (2.r-6)
dy
- бх +1
2(6)-6
2
(6)-6(6)+1
(6,0)
фу
d6,0
6
1
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фу 1 d x2-6x+1 (2.r-6) dy - бх +1 2(6)-6 2 (6)-6(6)+1 (6,0) фу d6,0 6 1

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Step 3

Thus, the equation of the line ...

(у-0) -6(х-6)
у%3 бх -36
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(у-0) -6(х-6) у%3 бх -36

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