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).  

Answered by Anish P. Maths tutor

2622 Views

See similar Maths A Level tutors

Related Maths A Level answers

All answers ▸

How do I prove (x-2) is a factor of the function f(x) = x^2-4x+4?


Differentiate: f(x)=2(sin(2x))^2 with respect to x, and evaluate as a single trigonometric function.


x = t^3 + t, y = t^2 +1, find dy/dx


Differentiate 3x^2 + 6x^5 + 2/x


We're here to help

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

© MyTutorWeb Ltd 2013–2024

Terms & Conditions|Privacy Policy
Cookie Preferences