# Derivatives of Logarithmic Functions.

I've been working through my practice problems and came across one that has stumped me.

$$y = (3x^{2}+2)^{ln x}$$

The answer to this is:

$$\frac{dy}{dx} = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2)+\frac{6xlnx}{3x^{2}+2})$$

What I'm coming up with is:

$$\frac{dy}{dx} = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2)+\frac{6x}{3x^{2}+2})$$

What I'm not understanding is where the $\frac{6xlnx}{3x^{2}+2}$ comes from, if anyone could explain this I'd really appreciate it.

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+1 for showing work, specifically noting where you want help, and for $\LaTeX$ use! :) –  anorton Mar 8 '13 at 4:10
Using logarithmic differentiation, that is, $$\ln(y)=\ln(x)\ln(3x^2+2).$$Now, when you take the derivative, the left-hand side becomes $$\frac{y'}{y}.$$ Use the product rule on the right-hand side, and don't forget $y=(3x^2+2)^{\ln(x)}.$ Thus, altogether you'll get $$\frac{y'}{y}=\frac{1}{x}\ln(3x^2+2)+\frac{\ln(x)}{3x^2+2}(6x).$$ Above, the $6x$ comes from the chain rule. Multiply both sides by $y=(3x^2+1)^{\ln(x)}$ and we get $$y'=(3x^2+2)^{\ln(x)}\left(\frac{1}{x}\ln(3x^2+2)+\frac{6x\ln(x)}{3x^2+2}\right).$$
We can perhaps go a little further in simplification to get $$y'=(3x^2+2)^{\ln(x)+1}\left(\frac{1}{x}+\frac{6x\ln(x)}{(3x^2+2)^2}\right).$$Here, I have factored a $(3x^2+2)$ from each term, thus the second term will get a square in the denominator, and the factor in front of the fraction gets an extra $+1$ in the numerator.
Oh I know, the way I came to this was: $$ln y = ln (3x^{2}+2)^{lnx}$$ $$\frac{1}{y} y' = \frac{1}{x} ln (3x^{2}+2) + \frac{1}{3x^{2}+2} 6x$$ $$y' = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2) + \frac{1}{3x^{2}+2} 6x)$$ $$y' = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2) + \frac{6x}{3x^{2}+2})$$ –  user1327636 Mar 8 '13 at 4:27