How do I differentiate y=x^x?

y=x^x

To find the differential, dy/dx, logs of both sides must be taken:

log(y)=log(x^x)

Then using log rules, the power can be brought down, outside the log expression:

log(y) = x log(x)

This expressions can now be differentiated with respect to x, using the chain rule on the left and the product rule on the right, giving:

(1/y) * dy/dx = 1 + log(x)

Multiplying through by y gives:

dy/dx = y (1 + log(x)) 

Remember! From the start of the question y=x^x, so this can be rewritten to:

dy/dx = x^x + x^xlog(x)