anonymous
  • anonymous
how do you derive the derivative of cotx? I know how to do sin, cos, csc, sec, tan
Mathematics
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
d/dx (cotx) = -csc^2(x)
anonymous
  • anonymous
how do you derive it? I know what it is.
bahrom7893
  • bahrom7893
use u sub: u = Sinx du = Cosx

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anonymous
  • anonymous
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)
bahrom7893
  • bahrom7893
rewrite Cot = Cos/Sin
bahrom7893
  • bahrom7893
please fan us both if we helped!
anonymous
  • anonymous
What identity does it become? Thats where i got confused in the quotient rule
bahrom7893
  • bahrom7893
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
anonymous
  • anonymous
its calc 1, I havent done integrals yet
bahrom7893
  • bahrom7893
WOOPS LOL sorry i misread the question
bahrom7893
  • bahrom7893
Use Cot = Cos*(Sin)^(-1) And use product rule
anonymous
  • anonymous
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)
bahrom7893
  • bahrom7893
Or Cot = Cos/Sin and quotient rule
anonymous
  • anonymous
beautiful, thanks pyeh9
anonymous
  • anonymous
my professor wanted us to derive all identities so I was confused on that one for the exam tomorrow
anonymous
  • anonymous
sure thing!

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