How do I solve [ln(x)]^(ln(x))?

I arrived at e^[ln(x)ln[lx(x)] . [1/x . ln[ln(x)]+ln(x).(1/(xln(x))]

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- Nov 16th 2013, 09:43 PMDarrylcwcLogarithmic differentiation
How do I solve [ln(x)]^(ln(x))?

I arrived at e^[ln(x)ln[lx(x)] . [1/x . ln[ln(x)]+ln(x).(1/(xln(x))] - Nov 16th 2013, 10:49 PMProve ItRe: Logarithmic differentiation
You can't SOLVE anything as you don't have an EQUATION.

However, if as your title suggests, you are trying to find the DERIVATIVE of , then start by taking the logarithm of both sides and simplify the RHS before differentiating. - Nov 16th 2013, 10:53 PMDarrylcwcRe: Logarithmic differentiation
That's what I did but the answer turned out very differently.

Could you demonstrate to to me?

How do I express e^[ln(x)ln(ln(x)] as a power of x? - Nov 16th 2013, 10:54 PMProve ItRe: Logarithmic differentiation
No, I want you to try it yourself. You need to post exactly what you tried, then we can see where you've gone wrong (if at all).

- Nov 16th 2013, 11:01 PMDarrylcwcRe: Logarithmic differentiation
I spent 4 hrs on this question. The reason why I don't like typing equations using keyboard is because it's slow and frustrating, especially when the equation becomes long and chaotic.

Simply put, I utilized ln, e and the chain rule to differentiate the RHS. Answer turned out very differently, however. Using the same mathematical reasoning, I have no issues with the other question. It's probably a matter of algebraic simplification-hence, I want the confirmation.

If that is true-a matter of simplication-then, I'll skip past the question as I never bother with the rudimentary concrete details.

I applied chain rule to y = e^[ln(x)ln(ln(x)] and I'm very sure I have to. If you could tell me if this part is correct, it is sufficient as the steps beyond the chain rules are tautological simplication. - Nov 16th 2013, 11:59 PMDarrylcwcRe: Logarithmic differentiation
Solved.