How do you differentiate (x^2)^(ln[x])?

In summary, to differentiate (x^2)^{ln x}, we can use the chain rule to write it as e^{2(ln x)^2} and then differentiate to get (4ln x)/x * e^{2(ln x)^2} or we can take the logarithm of both sides and differentiate to get (4ln x)/x * (x^2)^{ln x}.
  • #1
indie452
124
0

Homework Statement



differentiate (x2)lnx


im having a blonde moment...how do you start?
 
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  • #2
Write x as [tex]e^{\ln x}[/tex].
 
  • #3
ok so you mean write

eln(x2)^ln(x)

this gives me eln(x)*ln(x2)
 
  • #4
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.
 
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  • #5
Cyosis said:
can write the part in the exponent slightly easier by using [itex]\ln x^2=2\lnx[/itex].

how can you say this?
 
  • #6
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].
 
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  • #7
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
ok i got

e(2(lnx)2)

differentiating gives

(4lnx)/x * e(2(lnx)2)
 
  • #9
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].)
 

FAQ: How do you differentiate (x^2)^(ln[x])?

What is the formula for differentiating (x^2)^(ln[x])?

The formula for differentiating (x^2)^(ln[x]) is d/dx[(x^2)^(ln[x])] = (x^2)^(ln[x]) * (2ln[x] + 1/x).

What is the rule for differentiating a power raised to another power?

The rule for differentiating a power raised to another power is d/dx[(f(x))^n] = n(f(x))^(n-1) * f'(x).

Can the power rule be used to differentiate (x^2)^(ln[x])?

Yes, the power rule can be used to differentiate (x^2)^(ln[x]) as it follows the formula d/dx[(f(x))^n] = n(f(x))^(n-1) * f'(x).

What is the derivative of ln[x]?

The derivative of ln[x] is 1/x.

Is the derivative of (x^2)^(ln[x]) continuous?

Yes, the derivative of (x^2)^(ln[x]) is continuous as it follows the rules of differentiation and the function is continuous for all values of x.

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