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1 January, 14:47

If y=2ln (x2+2) y=2ln⁡ (x2+2), find the equation of the tangent line to the curve at x=4x=4.

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  1. 1 January, 17:23
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    Y=21n (x2+2) y=21n (x2+2)

    21n (x2+2) (y-1) = 0

    n=0, x=i/2-i/2, y=1
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