let y=x^x find dy/dx.... why is it hard ?

**Khonics89** · August 29th 2009, 01:11 AM

This question is from a UNIVERSITY PAPER (just incase some of you ppl will ask why would you want to differentiate that) ....

how would i go on about this... the answer is x^x(1+ln(x)) ???????

Thank you....

**Twig** · August 29th 2009, 01:36 AM

Hi!

Write $x^{x}$ as $e^{x\cdot \ln(x)}$ .

Now differentiate using normal rules, with $u(x)=x\cdot \ln(x)$ .

$\frac{d}{dx}(e^{u(x)})=e^{u}\cdot u'(x) = e^{x\cdot \ln(x)}\cdot(1+\ln(x) =x^{x}\cdot(1+\ln(x))$

Note that I used the product rule on $x\cdot \ln(x)$

**Rapha** · August 29th 2009, 01:37 AM

Hello Khonics89!

Originally Posted by Khonics89

This question is from a UNIVERSITY PAPER (just incase some of you ppl will ask why would you want to differentiate that) ....

how would i go on about this... the answer is x^x(1+ln(x)) ???????

Thank you....

Hint:

$\displaystyle x^{\displaystyle x} = e^{\displaystyle x\ln(x)}$

Do you know how to differentiate that?

Yours
Rapha

Edit: Beaten to it by Twig.