How do you differentiate x^x?

To differentiate xx, we first let y = xx. (Note that xx is not in the form xc where c is a constant or ax where a is a constant so the usual differentiation formulas cannot be used). The trick here is to take the natutral logarithm of both sides. Then you obtain, ln(y) = ln(xx). From here you need to use the rule that ln(xx) = xln(x). So currently we have ln(y) = xln(x). From here we can differentiate implicitly to get: 1/y multiplied by dy/dx = ln(x) + 1 (differentiate right hand side using product rule and left hand side using chain rule).The final step is to multiply through by y and substitute xx back in for y. This gives you: dy/dx = xx(ln(x) + 1).  

AP
Answered by Anish P. Maths tutor

3191 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

Use the substitution u=3+(x+4)^1/2 to find the integral of 1/(3+(x+4)^1/2) dx between 0 and 5.


How do I find the equation of the normal to the curve y=x^2 at the point (x1,y1)? Where x1=2 and y1=4 .


Find X log(x)=4 Base 10


Differentiate y = 7(x)^2 + cos(x)sin(x)


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2025 by IXL Learning