sorry to ask a second question so soon, but i'm just not getting this one. if f(x)= 3x lnx, then f'(x)=? i used f'(x)=3x(D lnx) + D (3x) (lnx) f'(x)=3x (1/x) + 3 (lnx) so... f'(x)=3+3lnx or 3(1+lnx). unfortunately that isn't one
Find lim x->1+ of [(1/(x-1))-(1/lnx)]. Here is my work... =(lnx-(x-1)) / ((x-1)(lnx)) =(lnx-1) / (lnx+ (x+1)/x) This becomes(1/x) / ((1/x)+(1/x^2)) which becomes 1/ (1/x^2) This equals 1/2. I understand the answer has to be
Find the derivative of y with respect to x. y=(x^6/6)(lnx)-(x^6/36) So far this is what I've gotten: y=(x^6/6)(lnx)-(x^6/36) y=(1/6)x^6(lnx)-(1/36)x^6 y'=(1/6)x^5(1/x)+lnx(x^5)-(1/6)x^5 What do I do now?