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I am presented with the following task:

"Find $f'(1)$ using logarithmic differentiation, given that

$$f(x) = \frac{(x^2-1)(x^2-2)(x^2-4)}{(x^2+1)(x^2+2)(x^2+3)}$$"

I have done the following:

$$\ln f(x) = \ln \frac{(x^2-1)(x^2-2)(x^2-4)}{(x^2+1)(x^2+2)(x^2+3)}$$ $$\ln f(x) = (\ln (x^2 -1)+\ln(x^2-2)+\ln(x^2-4))- (\ln(x^2+1)+\ln(x^2+2)+\ln(x^2+3))$$ $$\frac{f'(x)}{f(x)} = (\frac{2x}{x^2-1}+\frac{2x}{x^2-2}+\frac{2x}{x^2-4})-(\frac{2x}{x^2+1}+\frac{2x}{x^2+2}+\frac{2x}{x^2+3})$$

Wouldn't this leave a 0 in the denominator as soon as I insert $f'(1)$, seeing as $x^2-1$ remains a factor? I know that the solution is $\frac{1}{4}$ after checking with a digital resource.

Andrew Thompson
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    You haven't solved for $f'(x)$ yet. When you multiply both sides by $f(x)$, the terms that trouble you will cancel against similar terms appearing in $f(x)$. – vadim123 Oct 10 '13 at 16:29
  • Naturally, that allows me to leave out quite a lot, too, given 0 in the numerator. Thank you. – Andrew Thompson Oct 10 '13 at 16:43

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