Why is the deivative 2ln(x) is the same as that of ln(x^2)

[mahmoud2011]

In the textbooks some times they simplify the the logarithmic using logarithmic properties . But sometimes the domain of the simplified form is not the same as that of not simplified so how their derivative are equal. ??

 

[disregardthat]

The domains of the derivatives are not equal.

 

[mahmoud2011]

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

 

[disregardthat]

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

We don't have to, but we can. However note that 2log(x) is not a simplification of log(x^2) for negative x.

 

[mahmoud2011]

That is what I am talking about.

 

[SteveL27]

In the textbooks some times they simplify the the logarithmic using logarithmic properties . But sometimes the domain of the simplified form is not the same as that of not simplified so how their derivative are equal. ??

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

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

 

[uart]

That is what I am talking about.

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.

$$\frac{d}{dx} \left( 2 \log_e(x) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x>0$$

whereas

$$\frac{d}{dx} \left(\log_e(x^2) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x \neq 0$$

 

[SteveL27]

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.

$$\frac{d}{dx} \left( 2 \log_e(x) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x>0$$

whereas

$$\frac{d}{dx} \left(\log_e(x^2) \right) = \frac{2}{x} \,\,\,\,\,\,:\,\,\,\, x \neq 0$$

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.

 

[mahmoud2011]

Thanks for you all . Every time I use logarithmic laws I check the domains but my textbook doesn't mention that.

 

[Mark44]

If your textbook simplifies ln(x²) 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.

 

[mahmoud2011]

If your textbook simplifies ln(x²) 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.

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.