# Calculus, differentiation, but first taking ln on both sides?

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?

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

• 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 $\frac{d}{dx}(ln(f(x))=\frac{1}{f(x)}f'(x)$... – Sean Clark Nov 18 '10 at 15:51
• 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'! – Sean Clark Nov 18 '10 at 22:18