Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
+1 for showing work, specifically noting where you want help, and for $\LaTeX$ use! :) – apnorton Mar 8 '13 at 4:10
up vote 1 down vote accepted

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.

share|cite|improve this answer
According to my prof, that's the correct solution which I posted in the OP. – Matt Brzezinski Mar 8 '13 at 4:21
Oh I know, the way I came to this was: $$ln y = ln (3x^{2}+2)^{lnx}$$ $$\frac{1}{y} y' = \frac{1}{x} ln (3x^{2}+2) + \frac{1}{3x^{2}+2} 6x$$ $$y' = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2) + \frac{1}{3x^{2}+2} 6x)$$ $$y' = (3x^{2}+2)^{lnx} (\frac{1}{x}ln(3x^{2}+2) + \frac{6x}{3x^{2}+2})$$ – Matt Brzezinski Mar 8 '13 at 4:27
I've expanded my solution slightly and corrected some things. Hopefully it's clear. – Clayton Mar 8 '13 at 4:33

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.