f(x) = x^x, find f'(3).

Therefore, y = x^xcan then natural log both sides leaving ln(y) = xln(x) then differentiating both sides wrst to x d/dx(ln(y)=xln(x))we are then left with this expression (dy/dx)(1/y)=ln(x)+1 multiplying up by y leaves us with the expression dy/dx=y(ln(x)+1) can then substitue old expression back into new one and get this dy/dx=(x^x)(ln(x)+1) finally subbing in x=3 gives us f'(3)=27(ln(3)+1)