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

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d/dx (cotx) = -csc^2(x)
how do you derive it? I know what it is.
use u sub: u = Sinx du = Cosx

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Other answers:

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)
rewrite Cot = Cos/Sin
please fan us both if we helped!
What identity does it become? Thats where i got confused in the quotient rule
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
its calc 1, I havent done integrals yet
WOOPS LOL sorry i misread the question
Use Cot = Cos*(Sin)^(-1) And use product rule
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)
Or Cot = Cos/Sin and quotient rule
beautiful, thanks pyeh9
my professor wanted us to derive all identities so I was confused on that one for the exam tomorrow
sure thing!

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