P Pote New Member Joined Apr 1, 2005 Messages 10 Gender Female HSC 2006 Nov 6, 2005 #1 this is a very easy question, but i still need help -.- can anyone show me how to show: d [tan^-1 (cot x)] / dx = -1 i just learnt this, and my concepts still aren't that clear. thanks for your help!
this is a very easy question, but i still need help -.- can anyone show me how to show: d [tan^-1 (cot x)] / dx = -1 i just learnt this, and my concepts still aren't that clear. thanks for your help!
I icycloud Guest Nov 6, 2005 #2 y = arctan(cot(x)) Let u = cot(x) du = -csc2(x) dx Now y = arctan(u) dy = du / (1 + u2) = -csc2(x) dx / (1 + cot2(x)) Thus, dy/dx = -csc2(x)/csc2(x) = -1 #
y = arctan(cot(x)) Let u = cot(x) du = -csc2(x) dx Now y = arctan(u) dy = du / (1 + u2) = -csc2(x) dx / (1 + cot2(x)) Thus, dy/dx = -csc2(x)/csc2(x) = -1 #