how do you find the derivative of x^(ln(x))

Question

e.mccormick · Answer

Know the chain rule?

zoorie · Answer

yeah

e.mccormick · Answer

OK, and do you know implicit?

zoorie · Answer

yup

e.mccormick · Answer

Then you can use e or ln to change thing on both sides of the = and do implicit.  It will probably take a chain in there.

e.mccormick · Answer

Hmmmm.... looking at some reference.  $a^b=e^{b\ln a}$ therefore:

$x^{\ln(x)}=e^{\ln(x)\cdot\ln(x)}$

That should also help because now you can use the rules for e.

e.mccormick · Answer

So the implicit one would be:

$y=x^{\ln x} \implies$

$\ln(y)=\ln (x^{\ln x}) \implies$

$\ln y=\ln x\cdot \ln x $

which is what I was thinking about originally.  It would use implicit to solve, but the change of base method also looks useful.  So start with one of those and see what you get.

zoorie · Answer

how does $$\ln \left( x ^{\ln x} \right)=\ln x \times \ln x$$

e.mccormick · Answer

Basic log rules.  The power can jump out front.

$\log_b(x^a)=a \log_b(x)$

e.mccormick · Answer

The other two basic log rules can also be used to help simplify things in calculus:

$\log_n(ab) = \log_n(a)+\log_n(b)$

$\log_n\left(\dfrac{a}{b}\right) = \log_n(a)-\log_n(b)$

Calculus uses everything before it, all the way down to 1+1=2.  So you need to remember every algebra and trig rule you learned.

zoorie · Answer

ohhhhhhh