Emmanuel Pace

2022-07-20

$y=\left(3{x}^{2}+2{\right)}^{lnx}$
$\frac{dy}{dx}=\left(3{x}^{2}+2{\right)}^{lnx}\left(\frac{1}{x}ln\left(3{x}^{2}+2\right)+\frac{6xlnx}{3{x}^{2}+2}\right)$
What I'm coming up with is:
$\frac{dy}{dx}=\left(3{x}^{2}+2{\right)}^{lnx}\left(\frac{1}{x}ln\left(3{x}^{2}+2\right)+\frac{6x}{3{x}^{2}+2}\right)$
What I'm not understanding is where the $\frac{6xlnx}{3{x}^{2}+2}$ comes from, if anyone could explain this I'd really appreciate it.

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umgangistbf

Expert

Using logarithmic differentiation, that is,
$\mathrm{ln}\left(y\right)=\mathrm{ln}\left(x\right)\mathrm{ln}\left(3{x}^{2}+2\right).$
Now, when you take the derivative, the left-hand side becomes
$\frac{{y}^{\prime }}{y}.$
Use the product rule on the right-hand side, and don't forget $y=\left(3{x}^{2}+2{\right)}^{\mathrm{ln}\left(x\right)}.$ Thus, altogether you'll get
$\frac{{y}^{\prime }}{y}=\frac{1}{x}\mathrm{ln}\left(3{x}^{2}+2\right)+\frac{\mathrm{ln}\left(x\right)}{3{x}^{2}+2}\left(6x\right).$
Above, the 6x comes from the chain rule. Multiply both sides by $y=\left(3{x}^{2}+1{\right)}^{\mathrm{ln}\left(x\right)}$ and we get
${y}^{\prime }=\left(3{x}^{2}+2{\right)}^{\mathrm{ln}\left(x\right)}\left(\frac{1}{x}\mathrm{ln}\left(3{x}^{2}+2\right)+\frac{6x\mathrm{ln}\left(x\right)}{3{x}^{2}+2}\right).$
We can perhaps go a little further in simplification to get
${y}^{\prime }=\left(3{x}^{2}+2{\right)}^{\mathrm{ln}\left(x\right)+1}\left(\frac{1}{x}+\frac{6x\mathrm{ln}\left(x\right)}{\left(3{x}^{2}+2{\right)}^{2}}\right).$
Here, I have factored a $\left(3{x}^{2}+2\right)$ 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.

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