y ex lnxSolution Hope this is helpful i think that you shou

Solution

Hope this is helpful i think that you should use the quotient rule on the x/lnx part and find that derivative. It ends up being (lnx - 1)/(lnx)^2. <--- that is the derivative of the exponent. Y=e^u the derivative of e^x is e^x so that is easy. Y\'=e^u next you have to do u(x) u(x)= x/lnx u\'(x)= (lnx - 1)/(lnx)^2 e^u where u is u(x) so you get = e^x/lnx (e^x/lnx)(lnx - 1)/(lnx)^2 Answer: (e^x/lnx) x (lnx - 1)/(lnx)^2)