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Homework Help: Sin x vs sin^2 x

  1. Mar 15, 2007 #1
    1. The problem statement, all variables and given/known data
    Find any critical numbers of the function

    2. Relevant equations
    sin^2 x + cos x

    3. The attempt at a solution

    I actually have a sort of silly question. Woud sin^2 in the equation be solved using the differeniation rule of d/dx[sin x] = cos x or d/dx [sin u] = (cos u)d/dx u?
  2. jcsd
  3. Mar 15, 2007 #2
    d/dx (sin^2 x)= d/dx (sin x * sin x)
    = d/dx (sin x) * sin x + sin x * d/dx (sin x)
    =2 d/dx (sin x) sin x
    =2cos x sin x
  4. Mar 15, 2007 #3
    So I follow that thinking, but that would be the same as d/dx[sin u] right?
  5. Mar 15, 2007 #4
    No- I think you're referring to a case like this:

    A=d/dx sin(f(x))
    substitue u=f(x)

    A=du/dx * d/du sin(u)
    =[d/dx f(x) ] cos u
    =[d/dx f(x) ] cos(f(x))

    Remember, sin^2(x) does not equal sin (sin (x)) or sin (x^2)
    rather sin^2(x)= sin (x) * sin (x)
  6. Mar 15, 2007 #5
    Thank you.
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