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

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

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

Thank you....

2. 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 $x\cdot \ln(x)$

3. 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) ....

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.

4. ohh yes ofcourse hehe.. xD

Thanks guys.. never knew we can write x^x as e^xln(x)....

yeah all good thanks everyone

5. logarithmic differentiation ...

$y = x^x$

$\ln(y) = x \cdot \ln(x)$

$\frac{y'}{y} = 1 + \ln(x)$

$y' = y[1 + \ln(x)] = x^x[1 + \ln(x)]$