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How to Differentiate Cos^2 x?

  1. Aug 10, 2015 #1
    1. The problem statement, all variables and given/known data

    What is ##\frac{d}{dx}(cos^2 x)##?

    2. Relevant equations


    3. The attempt at a solution

    u = cos x
    ##\frac{du}{dx} = -\sin x##
    ##\frac{d}{dx}(cos^2 x) = \frac{d}{du}(u^2) \ \frac{du}{dx}##
    ##= 2u \ \frac{du}{dx}##
    ##= 2 \cos x (-\sin x)##
    ##= -2 \cos x \sin x##

    Is it correct?
     
  2. jcsd
  3. Aug 10, 2015 #2

    Mark44

    Staff: Mentor

    Yes, your work looks fine.
     
  4. Aug 10, 2015 #3

    Orodruin

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    Staff Emeritus
    Science Advisor
    Homework Helper
    Gold Member

    Yes. You may simplify it further using ##2\sin(a)\cos(a) = \sin(2a)##.
     
  5. Aug 10, 2015 #4
    That is correct
     
  6. Aug 10, 2015 #5
    Why the result is different when I integrate -2 cos x sin x back?

    ##\int -2 \cos x \sin x \ dx##

    u = sin x

    ##\frac{du}{dx} = \cos x##
    ##du = \cos x \ dx##

    ##\int -2 \cos x \sin x \ dx##
    ##= \int -2 \sin x (\cos x \ dx)##
    ##= \int -2u \ du##
    ##= -\frac{2}{1+1}u^{1+1} + c##
    ##= -\frac{2}{2} u^2 + c##
    ##= -u^2 + c##
    ##= -(\sin x)^2 + c##
    ##= - \sin^2 x + c##
     
  7. Aug 10, 2015 #6

    Mark44

    Staff: Mentor

    It's possible for ##\int f(x)## to be equal to ##\int g(x)dx##, even though f and g aren't the same. However, they can differ by most a constant. In your case ##\cos^2(x) = -\sin^2(x) + 1##. In other words, these two functions differ by 1.
     
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