Posted by bobby on Wednesday, March 28, 2012 at 9:15pm.
ln f = ln(x^6) + ln(x-8)^2 - ln(x^2+4)^7
= 6ln(x) + 2ln(x-8) - 7ln(x^2+4)
1/f f' = d/dx 6ln(x) + 2ln(x-8) - 7ln(x^2+4)
f ' = (6/x + 2/(x-8) - 14x/(x^2+4)) * (x^6 * (x-8)^2) / (x^2+4)^7
you can massage that in various ways. One way ends up as
-2(x-8)(x^5)(3x^3-32x^2-16x+96)/(x^2+4)^8
plug in x=8 to get the value. That (x-8) in the numerator makes it all 0.
Actually, if you hadn't been asked for f '(x), you could have saved all the work by noting that if (x-k)^n is a factor of f(x), and n > 1, then (x-k) will be a factor of each term of f '(x), so f '(k) will be zero.
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