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$ y = \ln(5x^2 + 9)^3$ Find derivative.

So my initial answer for this was to bring the $3$ to the front of $ln$ to get:
$3 \ln (5x^2 + 9)$ and from there use the the properties of ${d\over dx}\ln x={1\over x}$ and derivative of $5x^2+9$ to get: ${30x \over 5x^2 +9}$.

But the answer provided on my assignment includes a ${1 \over (5x^2+9)^3} \times \cdots$ Which I'm assuming means that the equation was meant to be read as $y = \ln[(5x^2 + 9)^3]$.

So am I just wrong of can the question be interpreted different ways?

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  • $\begingroup$ You realise that if you brought the 3 to the front you had to have been interpreting it as $y=ln[(5x^2+9)^3]$? $\endgroup$ – nathan.j.mcdougall Nov 2 '14 at 2:47
  • $\begingroup$ Not if it was the result of the chain rule. $\endgroup$ – Matt Samuel Nov 2 '14 at 2:49
  • $\begingroup$ Ah, I see what he means, I thought he was using $ln{a^b}=blna$ $\endgroup$ – nathan.j.mcdougall Nov 2 '14 at 2:50
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If $$y=\log[(5x^2+9)^3]$$ then your answer $$y'==\frac{30x}{(5x^2+9)}$$ is perfectly correct.

However, it the problem is $$y=\log^3[(5x^2+9)]$$ the story is totally different and differentiation will lead to $$y'=\frac{30 x \log ^2\left(5 x^2+9\right)}{5 x^2+9}$$ But I do not see how could arrive here the $${1 \over (5x^2+9)^3} \times \cdots$$ you mention.

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In other way

$$y'= \frac{3(5x^2+9)^2\cdot10x}{(5x^2+9)^3}=\frac{30x}{(5x^2+9)}$$

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$y=ln(5x^2+9)^3$ is equivalent to $y=ln[(5x^2+9)^3]$. Generally, $y=[ln(5x^2+9)]^3$ is notated as $y=ln^3(5x^2+9)$. So there is only one possible interpretation.

As for the solution, $e^\frac{y}{3}=5x^2+9$, so implicitly, $\frac{1}{3}\frac{dy}{dx}e^\frac{y}{3}=10x$ and then $$\frac{dy}{dx}=\frac{30x}{5x^2+9}$$ Of course, the chain rule can also apply.

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