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How do i differentiate x^lnx

the answer is supposed to be
2ln(x)x^lnx-1
Reply 1
Avatar for noobynoo
noobynoo
6
xlnx=exp(ln(x)ln(x))=exp((ln(x))2)x^{ln x} = exp( ln(x)ln(x)) = exp( (ln(x))^2 )

so...

ddxexp((ln(x))2)=2ln(x)xexp((ln(x))2)=2ln(x)xln(x)−1\frac{d}{dx} exp( (ln(x))^2 ) = 2\frac{ln(x)}{x}exp( (ln(x))^2 ) = 2 ln(x)x^{ln(x)-1}
(edited 12 years ago)
Reply 2
Avatar for Gemini92
Gemini92
Original post by adibasa14
the answer is supposed to be
2ln(x)x^lnx-1


I'm not sure how to do this but have a look at http://www.wolframalpha.com which is really good for this stuff. Just put your equation in and where it says derivative, click show steps and it will show you how to differentiate your equation.
Reply 3
Avatar for Bjoorn
Bjoorn
11
Here is another way of going about this problem in case you haven't done that kind of differentiation.
y=xlnx y = x^{ln x}
y=elnx(lnx) y = e^{lnx^{(ln x)}}
Unparseable latex formula:

y = e^{(ln x) (ln x)}}


Now using the product rule y=eu y = e^{u} you can show easily that dudx=2lnxx\frac {du}{dx} = \frac {2lnx}{x}.
Thus dydx=dudxeu \frac {dy}{dx} = \frac {du}{dx}e^{u}
dydx=2lnxxe(ln2x) \frac {dy}{dx} = \frac {2lnx}{x}e^{(ln^{2} x)} (which is the same as the answers above just in a different guise.)
Remember that e(ln2x)e^{(ln^{2} x)} is the same as y=xlnx y = x^{ln x} if you want the other form

edit: just saw that noobynoo used the same method, but my presentation is neater if that makes any difference hehe
(edited 12 years ago)
Reply 4
Avatar for adibasa14
adibasa14
OP
thanks everybody :smile:

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