Question

how do you derive the derivative of cotx? I know how to do sin, cos, csc, sec, tan

Answer 1

d/dx (cotx) = -csc^2(x)

Answer 2

how do you derive it?
I know what it is.

Answer 3

use u sub:
u = Sinx
du = Cosx

Answer 4

oh. My mistake. Well, one way would be to use the quotient rule. So cot(x) = cos(x)/sin(x). Taking the derivative of cos(x)/sin(x) using the quotient rule yields:
[sin(x)[d/dx(cos(x)] - cos(x)[d/dx(sin(x))]]/sin^2(x)

Answer 5

rewrite Cot = Cos/Sin

Answer 6

please fan us both if we helped!

Answer 7

What identity does it become? Thats where i got confused in the quotient rule

Answer 8

Okay it becomes:
Integral Cot = Integral Cos/Sin
Let u = sin
du = Cos dx
Integral Cot = Integral du/u = Ln|u| + C = Ln|Sin(x)|+C

Answer 9

its calc 1, I havent done integrals yet

Answer 10

WOOPS LOL sorry i misread the question

Answer 11

Use Cot = Cos*(Sin)^(-1) And use product rule

Answer 12

from where I left off:
the top becomes -sin^2(x) - cos^2(x) = -1
then the whole thing becomes -1/sin^2(x) = -csc^2(x)

Answer 13

Or Cot = Cos/Sin and quotient rule

Answer 14

beautiful, thanks pyeh9

Answer 15

my professor wanted us to derive all identities so I was confused on that one for the exam tomorrow

Answer 16

sure thing!