solve for x:
(x^x^2) * (2x * (ln(x)) + x) = 0

Please show steps, thanks!
solve for x:
(x^x^2) * (2x * (ln(x)) + x) = 0

Please show steps, thanks!
@Mathematics

Question

solve for x:
(x^x^2) * (2x * (ln(x)) + x) = 0

Please show steps, thanks!
 solve for x:
(x^x^2) * (2x * (ln(x)) + x) = 0

Please show steps, thanks!
 @Mathematics

turingtest · Answer

$$x^{x^2}  (2x  \ln(x) + x) = 0$$ right?

anonymous · Answer

yeah that's right

anonymous · Answer

it's the derivative of $$x^{x^{2}}$$

anonymous · Answer

trying to find the interval which the derivative is >0 and >0. Having some trouble solving this though.

turingtest · Answer

I see Wolfram has the answer$$ x={1 \over \sqrt e}$$but it cannot show steps, I will continue to work on it and get back to you if I have any progress.

anonymous · Answer

ok, thanks

turingtest · Answer

Trying to find an interval which is what? Has derivative >0?

anonymous · Answer

trying to find the intervals where the derivative is greater than 0 and less than 0

anonymous · Answer

ok I've got it, you just multiply both sides by the (x^(x^2)) and just solve 2x * (ln(x)) + x = 0