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

[indie452]

Homework Statement

differentiate (x²)^lnx


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

 

[Cyosis]

Write x as $$e^{\ln x}$$.

 

[indie452]

ok so you mean write

e^{ln(x2)^ln(x)}

this gives me e^ln(x)*ln(x2)

 

[Cyosis]

Yes that's correct. You can write the part in the exponent slightly easier by using $\ln x^2=2\ln x$. Now you just have to use the chain rule.

 

[indie452]

can write the part in the exponent slightly easier by using $\ln x^2=2\lnx$.

how can you say this?

 

[Cyosis]

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. $$(x^2)^{\ln x}=(x)^{2 \ln x}=(e^{\ln x})^{\ln x^2}=e^{\ln x \ln x^2} \neq e^{(\ln x^2)^{\ln x}}$$.

 

[HallsofIvy]

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]

 

[indie452]

ok i got

e^(2(lnx)2)

differentiating gives

(4lnx)/x * e^(2(lnx)2)

 

[HallsofIvy]

Yes, that's right.

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

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

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