Find the derivative of x^x

y = x^x -- Yikes! Doesn't that look ugly. 
It's mostly the x^x part, since ... to the x is fine, and x to the ... is fine.
We must split the two x's in order to continue. 
Perhaps we could log? since then we can pull the index to the front. 
so: log(y) = log(x^x) = x log(x) -- because [log(a^b) = b log(a)]
Well this looks much better. On the left we have something which is easy using chain rule
and the right side, looks easy using product rule.

So:
d/dx log(y) = dy/dx * (1/y) 
and:
d/dx xlog(x) = 1log(x) + x*(1/x)
= log(x) + 1

so:
dy/dx * (1/y) = 1 + log(x)
Multiplying through by y, gives us:
dy/dx = y ( 1 + log(x) )
but remember, y = x^x 
so dy/dx = x^x ( 1 + log(x) )

Answered by Mayur D. Maths tutor

4412 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

You are given the equation y=x^2. Determine whether or not the equation has any maximums or minimums and identify them (whether they are maximums or minimums).


Integrate Sin(2X)


Integrate x*ln(x) with respect to x


How do you integrate the natural logarithm ln(x)?


We're here to help

contact us iconContact usWhatsapp logoMessage us on Whatsapptelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo
Cookie Preferences