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Quotient Rule for Derivative

  1. Feb 2, 2013 #1
    1. The problem statement, all variables and given/known data


    Use quotient rule
    2. Relevant equations

    d/dx (cot(x)

    3. The attempt at a solution
    d/dx (cot(x))
    = cos(x)/sin(x)
    = (sin(x))(-sin(x)) - (cos(x))(cos(x)) / (sin(x))^2
    = -(sin(x))^2 - (cos(x))^2 / (sin(x))^2
    = -1 - (cos(x))^2


    Does the last step equal 1 + (cos(x))^2 and how come that turns into negative -(csc(x))^2
     
  2. jcsd
  3. Feb 2, 2013 #2

    jbunniii

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    I think you dropped some parentheses and this led you to a wrong answer. That first step should be:
    $$ \frac{d}{dx} \cot(x) = \frac{\sin(x)(-\sin(x)) - \cos(x)\cos(x)}{\sin^2(x)}$$
    Now simplify the numerator and see what you end up with.
     
  4. Feb 2, 2013 #3
    = (-1)-(cos(x))^2
     
  5. Feb 2, 2013 #4

    jbunniii

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    Shouldn't the second term be divided by ##\sin^2(x)##?
     
  6. Feb 2, 2013 #5
    Hold on
    -1 - (cos(x))^2 / (sin(x))^2 = -1/(sin(x))^2 = -(csc(x))^2

    but...

    -(sin(x))^2 - (cos(x))^2 / (sin(x))^2 = 1/(sin(x))^2 = -(csc(x))^2
     
  7. Feb 2, 2013 #6

    jbunniii

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    How did you get that?
    I think there are some parentheses missing again. The expression on the left should be
    $$\frac{-\sin^2(x) - \cos^2(x)}{\sin^2(x)}$$
    Now you can use the identity ##\sin^2(x) + \cos^2(x) = 1## in the numerator.
     
  8. Feb 2, 2013 #7


    My main question is do both of the following equations equal -(csc(x))^2

    1/(sin(x))^2 and -1/(sin(x))^2

    or which one of those two define -(csc(x))^2 ?
     
  9. Feb 2, 2013 #8

    HallsofIvy

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    Since, by definition, csc(x)= 1/sin(x), it follows that sin(x)= 1/csc(x) and then that [itex]sin^2(x)= 1/csc^2(x)[/itex].

    If you want "-" on the right, you will have have a "-" on the right.
     
  10. Feb 2, 2013 #9

    Mark44

    Staff: Mentor

    You haven't actually taken the derivative yet, so the above should be:
    =d/dx(cos(x)/sin(x))

    Others have already commented about the missing parentheses and the incorrect parts.

     
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