Original equation:

$g(x) = \frac{(x+1)(x^2+2)(x^3+3)}{\sqrt{x^4+3}}$

If I take ln on both sides, and than differentiate I get this:

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

Is this good so far? I don't know what to do about the left side?

  • 2
    $\begingroup$ The derivative of $\log g(x)$ isn't $1/g(x)$. $\endgroup$ – Robin Chapman Nov 18 '10 at 15:34
  • $\begingroup$ Explains why wolframs solution is a lot longer :p $\endgroup$ – Algific Nov 18 '10 at 18:27

Close, but you forgot to apply the chain rule!

  • $\begingroup$ Were? :( left side? g(x)? $\endgroup$ – Algific Nov 18 '10 at 15:32
  • 1
    $\begingroup$ Both sides actually. It might help to write it out carefully: $ln(g(x))=ln(x+1)+ln(x^2+2)+ln(x^3+3)-\frac{1}{2}ln(x^4+3)$. But remember, the chain rule says that $\frac{d}{dx}(ln(f(x))=\frac{1}{f(x)}f'(x)$... $\endgroup$ – Sean Clark Nov 18 '10 at 15:51
  • $\begingroup$ But how do you actually apply that to the left part? I mean to ln(g(x)). Would you use the entire right side as f(x) after I've taken the derivative? $\endgroup$ – Algific Nov 18 '10 at 18:55
  • $\begingroup$ That is the left part, once you replace f with g. In symbols, the resulting equation is: $\frac{g'(x)}{g(x)}=\frac{d}{dx}\left(ln(x+1)+ln(x^2+2)+ln(x^3+3)−\frac{1}{2}ln(x^4+3)\right)$. The derivatives on the right side you can compute directly with the chain rule, and the final step is to multiply through by g(x) to isolate g'! $\endgroup$ – Sean Clark Nov 18 '10 at 22:18

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