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Differentiate (x^2)^(ln[x])

  • Thread starter indie452
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  • #1
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Homework Statement



differentiate (x2)lnx


im having a blonde moment...how do you start?
 

Answers and Replies

  • #2
Cyosis
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Write x as [tex]e^{\ln x}[/tex].
 
  • #3
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ok so you mean write

eln(x2)^ln(x)

this gives me eln(x)*ln(x2)
 
  • #4
Cyosis
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Yes that's correct. You can write the part in the exponent slightly easier by using [itex]\ln x^2=2\ln x[/itex]. Now you just have to use the chain rule.
 
Last edited:
  • #5
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can write the part in the exponent slightly easier by using [itex]\ln x^2=2\lnx[/itex].
how can you say this?
 
  • #6
Cyosis
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I 'said' that, because in the latex code I wrote 2\lnx, \lnx is not a command in latex so it doesn't recognize it. I added the space now so it becomes 2\ln x, which should display the correct result.

Edit: I see you edited post 3, but what you did there is not correct. [tex](x^2)^{\ln x}=(x)^{2 \ln x}=(e^{\ln x})^{\ln x^2}=e^{\ln x \ln x^2} \neq e^{(\ln x^2)^{\ln x}}[/tex].
 
Last edited:
  • #7
HallsofIvy
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He didn't say it. (Unless he editted his post immediately after your response.) He said ln(x2)= 2ln(x).

On edit: And while I was typing this, he explained!

Using that
[tex](x^2)^{ln(x)}= e^{(ln(x))(2ln(x))}= e^{2(ln(x))^2}[/itex]
 
  • #8
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ok i got

e(2(lnx)2)

differentiating gives

(4lnx)/x * e(2(lnx)2)
 
  • #9
HallsofIvy
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Yes, that's right.

Another way to differentiate [itex]y= (x^2)^{ln x}[/itex] is to take the logarithm of both sides: [itex]ln(y)= ln((x^2)^{ln(x)}= ln(x)(ln(x)^2)= 2 (ln(x))^2[/itex]

Now, differentiating both sides with respect to x,
[tex]\frac{1}{y}y'= 4 ln(x)\frac{1}{x}[/tex]
[tex]y'= \frac{4 ln(x)}{x} y= \frac{4 ln(x)}{x}(x^2)^{ln(x)}[/tex]

(If you were given the problem as [itex](x^2)^{ln x}[/itex] it is probably better to write the answer using that rather than [itex]e^{2(ln x)^2}[/itex].)
 

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