Emmanuel Pace

2022-07-20

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

The answer to this is:

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

What I'm coming up with is:

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

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.

The answer to this is:

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

What I'm coming up with is:

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

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.

umgangistbf

Beginner2022-07-21Added 12 answers

Using logarithmic differentiation, that is,

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

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 $y=(3{x}^{2}+2{)}^{\mathrm{ln}(x)}.$ Thus, altogether you'll get

$\frac{{y}^{\prime}}{y}=\frac{1}{x}\mathrm{ln}(3{x}^{2}+2)+\frac{\mathrm{ln}(x)}{3{x}^{2}+2}(6x).$

Above, the 6x comes from the chain rule. Multiply both sides by $y=(3{x}^{2}+1{)}^{\mathrm{ln}(x)}$ and we get

${y}^{\prime}=(3{x}^{2}+2{)}^{\mathrm{ln}(x)}(\frac{1}{x}\mathrm{ln}(3{x}^{2}+2)+\frac{6x\mathrm{ln}(x)}{3{x}^{2}+2}).$

We can perhaps go a little further in simplification to get

${y}^{\prime}=(3{x}^{2}+2{)}^{\mathrm{ln}(x)+1}(\frac{1}{x}+\frac{6x\mathrm{ln}(x)}{(3{x}^{2}+2{)}^{2}}).$

Here, I have factored a $(3{x}^{2}+2)$ 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.

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

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 $y=(3{x}^{2}+2{)}^{\mathrm{ln}(x)}.$ Thus, altogether you'll get

$\frac{{y}^{\prime}}{y}=\frac{1}{x}\mathrm{ln}(3{x}^{2}+2)+\frac{\mathrm{ln}(x)}{3{x}^{2}+2}(6x).$

Above, the 6x comes from the chain rule. Multiply both sides by $y=(3{x}^{2}+1{)}^{\mathrm{ln}(x)}$ and we get

${y}^{\prime}=(3{x}^{2}+2{)}^{\mathrm{ln}(x)}(\frac{1}{x}\mathrm{ln}(3{x}^{2}+2)+\frac{6x\mathrm{ln}(x)}{3{x}^{2}+2}).$

We can perhaps go a little further in simplification to get

${y}^{\prime}=(3{x}^{2}+2{)}^{\mathrm{ln}(x)+1}(\frac{1}{x}+\frac{6x\mathrm{ln}(x)}{(3{x}^{2}+2{)}^{2}}).$

Here, I have factored a $(3{x}^{2}+2)$ 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.

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