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- Thread starter mahmoud2011
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disregardthat

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The domains of the derivatives are not equal.

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I know but mustn't we simplify and take the derivative ,right ?

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disregardthat

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We don't have to, but we can. However note that 2log(x) is not a simplification of log(x^2) for negative x.I know but mustn't we simplify and take the derivative ,right ?

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That is what I am talking about.

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They are the same. The derivative of 2*ln(x) = 2/x.

The derivative of ln(x^2) = (1/x^2) * 2x = 2/x.

- #7

uart

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Yep I can see that Mahmoud.That is what I am talking about.

Yes you make a good point. I sometimes see them treated as being identical when as you point out they are not.

[tex]\frac{d}{dx} \left( 2 \log_e(x) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x>0[/tex]

whereas

[tex]\frac{d}{dx} \left(\log_e(x^2) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x \neq 0[/tex]

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Isn't the point just that the original functions 2log(x) and log(x^2) have different domains? This question really has nothing to do with derivatives. But it's a good point ... you can't arbitrarily apply the log laws without double-checking the domains.Yep I can see that Mahmoud.

Yes you make a good point. I sometimes see them treated as being identical when as you point out they are not.

[tex]\frac{d}{dx} \left( 2 \log_e(x) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x>0[/tex]

whereas

[tex]\frac{d}{dx} \left(\log_e(x^2) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x \neq 0[/tex]

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Mark44

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No , this problem doesn't exist but I wanted to say that why the textbook didn't mention that I must check the domains , Is the author wants me to do it myself and actually I did.^{2}) to 2 ln(x), they are probably making the assumption that x > 0. If they don't show that assumption anywhere, then they are being very sloppy.

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