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

what is the limit as h approaches 0 of ((x+h)^pi-x^pi)/h? Multiple choice answers: A: lnx B: lnx (x^pi) C: Pi lnx D: (pi^2) ln (x^2) E: (pi^x) lnx I know this is d/dx x^pi, or Pi x^(Pi-1)

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?

Find the derivatives, dy/dx for the following functions. a. x^2y^4 = lnx + 3y b. y = ln(2x^2 + 1) c. y = ln(ãx) d. y = lnx/x e. y = x^2 lnx f. y ß log_2(3x)