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In finding the derivative of $\ln(2x^2),\;$ I have applied the chain rule and obtained $2x / x^2.\;$ Is this correct?

If not could some please explain how to do it?

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    $\begingroup$ That is correct, but you can simplify it further to $\dfrac{2}{x}$. $\endgroup$ – Daniel Fischer Jan 16 '14 at 12:25
  • $\begingroup$ @DanielFischer Thank you! $\endgroup$ – Lauren Jan 16 '14 at 12:27
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If you use rules of logarithms, you don't even have to appeal to the chain rule.

Notice that $\ln\left(2x^{2}\right)=\ln\left(2\right)+2\ln\left(x\right)$, so the derivative is $\frac{2}{x}$.

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Another possible way is: if $x\neq 0$, then by setting $y=\ln(2x^2)$ we have $\exp(y)=2x^2$ so $y'e^y= (e^y)'=4x$ so $$y'=4xe^{-y}=4x\times\left(\frac{1}{2x^2}\right)=\frac{2}{x}$$

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    $\begingroup$ A simpler way is to use properties of logarithm, $\ln(2x^2) = \ln(2) + 2\ln(x)$, and take advantage of the derivative of a constant being zero. $\endgroup$ – hardmath Jan 16 '14 at 12:37
  • $\begingroup$ @hardmath: Honestly yes you are right. I made the solution difficult for the OP. :-) $\endgroup$ – mrs Jan 16 '14 at 12:38
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    $\begingroup$ I think it's great to take the time to show different routes leading to the same Answer. Priority has to be given to finding one way to do it you can remember for the test, and secondarily to doing it the fastest way. $\endgroup$ – hardmath Jan 16 '14 at 12:41
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Using the chain rule, if you have a function written as $h(x) = f(g(x))$, its derivative is $f'(g(x))g'(x)$. For $h(x)=\ln(2x^2)$, you can take $f$ and $g$ to be $$ f(x) = \ln(x)\\ g(x) = 2x^2$$

Deriving them gives $f'(x) = 1/x$ and $g'(x) = 4x$. Insert this into the chain rule and you get $$ h'(x) = \frac{1}{2x^2}\cdot 4x = \frac{2}{x} $$

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You did just fine, and you correctly applied the chain rule. You correctly found the derivative. All you may want to do now is to simplify $$\require{cancel}\Big(\ln(2x^2)\Big)' = \frac {2x}{x^2} = \frac {2\cancel{x}}{x\cdot \cancel{x}} = \dfrac 2{x}$$

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  • $\begingroup$ Do you think your answer contributes with anything that haven't been mentioned already? $\endgroup$ – Stefan Hansen Jan 16 '14 at 13:28
  • $\begingroup$ @Stafan Yes, I do. The OP asked if s/he applied the chain rule correctly, and s/he did. So "teaching" how to apply the chain rule is unnecessary (one answer), and the OP did not ask for other ways to take the derivative (two answers): s/he asked simply if his/her answer was correct, and if not, what is the correct answer. I answered by confirming that the OP's answer is correct, his/her application of the chain rule is correct, and pointed out that the only thing missing from the final answer was simply a matter of simplifying. $\endgroup$ – Namaste Jan 16 '14 at 13:36

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