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The problem I have is to differentiate $ y = ln(x^4)$

Using the rule : $$\frac{d[lnf(x)]}{dx}=\frac{f'(x)}{f(x)}$$ My working is: $$\frac{dy}{dx} = \frac{x^4ln(x)}{x^4}$$ $$=ln(x)$$

but the book is giving the answer : $$\frac{4}{x}$$

Please help because I am confused

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Caution: $$\log x^4=4\log |x|\neq x^4\log x\;\;!$$ –  DonAntonio Nov 27 '12 at 21:41
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2 Answers

$f(x) = x^4$. Thus $f'(x) = 4x^3$. Plug into provided rule.

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Thank you, I see it now –  user866190 Nov 27 '12 at 21:51
    
Note that the "rule" is merely an application of the chain rule. –  Jacob Nov 27 '12 at 21:55
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Following my comment and assuming $\,x>0\,$ to avoid the absolute value:

$$y=\log x^4=4\log x\Longrightarrow y'=\frac{4}{x}= $$

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Thank you, another way which makes it easy to work out... –  user866190 Nov 27 '12 at 21:53
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