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$y=\ln(x)^2$

I am not sure why the answer would be $\frac{2\ln(x)}{x}$

I used this property "power rule" "$\ln(x^n) = n\ln(x)$

So i got $2\ln(x) $

the derivative of that using the constant multiplier rule i got

$\frac{2}{x}$

can I use the other chain rule to $y=f(u)$ and $g=g(x)$ Am i not supposed to bring that 2 in front becuase the whole expression is getting raised not the $x$? Any help would be great,

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    $\begingroup$ Look at the parentheses, $(\log x)^2 \neq \log (x^2)$. $\endgroup$
    – Daniel Fischer
    Commented Oct 26, 2013 at 20:49
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    $\begingroup$ Careful: $$\log(x^n)=n\log x\neq (\log x)^n=\log^nx$$ $\endgroup$
    – DonAntonio
    Commented Oct 26, 2013 at 20:49

1 Answer 1

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Apply the chain rule with $\;f(x):=x^2\;,\;\;g(x):=\log x$ :

$$(f(g(x)))'=f'(g(x))\cdot g'(x)$$

So

$$(\log x)^2=2\log x\cdot \frac1x=\frac{2\log x}x$$

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  • $\begingroup$ So another way of looking at the chain rule is using the power rule multipilying the derivative of the inside function? i think that works? am i right? $\endgroup$
    – John
    Commented Oct 26, 2013 at 20:58
  • $\begingroup$ Well...yes, that's what I wrote, if I understood your comment correctly. $\endgroup$
    – DonAntonio
    Commented Oct 26, 2013 at 20:58
  • $\begingroup$ Yes the chain rule composition formula is confusing for me to look at thank you for the help $\endgroup$
    – John
    Commented Oct 26, 2013 at 21:00
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    $\begingroup$ John, if you accept somebody's answer, please give him an upvote. Particularly if the answer is correct. $\endgroup$
    – imranfat
    Commented Oct 26, 2013 at 22:06

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