This is a question from my maths book:

Take Logarithms of both sides and use the log laws to differentiate

(a) y=x^x

for this I did:

log y =x log x

then I do not know how to go on from there. I know that the derivative of log x is 1/x

Take ln of both sides:

ln(y)=ln(x^{x})

Property of logs:

ln(y)=xln(x)

Differentiate:

\frac{y'}{y}=ln(x)+1

Solve for y':

y'=y(ln(x)+1)

Remember that y=x^{x} and sub in:

y'=x^{x}(ln(x)+1)