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Simple differentiation

  1. Aug 5, 2010 #1
    Can someone show me why the derivative of

    k(x) = cot x

    Is

    k'(x) = -cosec^2 x

    And not

    k'(x) = -1/tan^2

    Thanks!
     
  2. jcsd
  3. Aug 5, 2010 #2

    Char. Limit

    User Avatar
    Gold Member

    Well, cot(x) is really the quotient of two functions: cos(x)/sin(x). Then, using the quotient rule...

    [tex]\frac{d cot(x)}{dx} = \frac{(sin(x))(-sin(x)) - (cos(x))(cos(x))}{sin^2(x)}[/tex]

    [tex]=\frac{-sin^2(x)-cos^2(x)}{sin^2(x)}[/tex]

    [tex]=-\frac{sin^2(x)+cos^2(x)}{sin^2(x)}[/tex]

    [tex]=-\frac{1}{sin^2(x)}[/tex]
     
  4. Aug 5, 2010 #3
    One way to derive the formula for the derivative of cot(x) is to rewrite it as 1/tan(x) (or cos(x)/sin(x)) and use the quotient rule. The mistake you probably made in getting -1/tan^2(x) was to treat cot(x) as the composition of the functions y=1/u(x) and u(x)=tan(x) and then forgetting to apply the chain rule: you must multiply -1/tan^2(x) by the derivative of tan(x).
     
  5. Aug 5, 2010 #4
    Thanks to you both. That's very clear.
     
  6. Aug 5, 2010 #5
    Just for information, the mistaken thought process I had was to differentiate tan^-1 as -1* tan^-2....
     
  7. Aug 5, 2010 #6

    Mute

    User Avatar
    Homework Helper

    You can do that, but your calculation is incomplete. You have to use the chain rule:

    [tex]\frac{d}{dx}(f(g(x))) = \frac{df(g)}{dg} \frac{dg(x)}{dx}[/tex]

    With f(g) = 1/g and g = tan(x), you should again find that the derivative of cot(x) is -csc(x).
     
  8. Aug 5, 2010 #7

    Mark44

    Staff: Mentor

    As Mute pointed out, this will work if you use the chain rule correctly. However, the expression tan-1(x) usually is taken to mean the inverse of the tangent function (AKA the arctangent), not its reciprocal.

    If you write d/dx(cot(x)) = d/dx(tan-1(x)), you are really going to confuse a lot of people.
     
  9. Aug 6, 2010 #8
    Thankyou.
     
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