x^x and sin(x)/x The derivative of x^x is: y=x^x ln(y)=xlnx (1/y)y'=lnx+x*1/x y'=y(lnx+1) y'=x^x(lnx+1) Why is this not a possible way of solving it: y=x^x y=(e^lnx)^x y=e^(xlnx) y'=e^(xlnx)lnx y'=x^x(lnx) I see it's not the same answer but I wonder why The reason is if the above was correct it would be easy to integrate x^x (which is not possible): y=x^x y=e^(xlnx) integral(y)=e^(xlnx)/lnx integral(y)=x^x/lnx why is this not correct? Ok I just realized that the inner derivative (do you also call it this?) of ln x is 1 thus (e^(xlnx))'=e^(xlnx)(lnx + (inner derivative of lnx)) which is one thus x^x(lnx+1) How does this affect the integral? Secondly how would you go about drawing sin(x)/x by hand without a calculator What I realize is the following: Though not defined x=0 would seem to be at 1 mainly since for small sin(x)/x=x thus 1/x*sin(x)=1/x*x)=1 Next the function would reach the first zero (at the positive side) at pi/4 so how would the line look between 0 and pi/4? I have seen it but I don't know how to realize it myself. Next I understand for x=pi/4->pi/2 since: -__*\_/ it would look something like Code (Text): \ / |_/ as opposed to \ / \_/ slightly bulged to the left due to decreasing 1/x but where between pi/4 and pi/2 would the maximum be? Following this the amplitude of the waves would just decrease which can be done by appreciation... Any ideas?