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Math Help - Trig

  1. #1
    Member srirahulan's Avatar
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    Lightbulb Trig

    If f(x)= cos^2x+cos^2(\frac{\pi}{3}+x)-cosxcos( \frac{\pi}{3}+x) Find out the f'(x). In this sum How can I Easily derivate By substituting Method like a=\frac{\pi}{3}
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  2. #2
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    Re: Trig

    So you would like to find out the following

    d/dx(cos^2(x)+cos^2(pi/3+x)-cos(x) cos(pi/3+x))

    Thus, Rewrite the expression as: cos^2(x)+cos^2(pi/3+x)-cos(x) cos(pi/3+x) = cos^2(x)-cos(x) sin(pi/6-x)+sin^2(pi/6-x)

    = d/dx(cos^2(x)-cos(x) sin(pi/6-x)+sin^2(pi/6-x))

    Differentiate the sum term by term and factor out constants:

    = d/dx(cos^2(x))-d/dx(cos(x) sin(pi/6-x))+d/dx(sin^2(pi/6-x))

    Using the chain rule, d/dx(cos^2(x)) = ( du^2)/( du) ( du)/( dx), where u = cos(x) and ( d)/( du)(u^2) = 2 u

    = -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))+2 cos(x) d/dx(cos(x))

    The derivative of cos(x) is -sin(x):

    = -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))+-sin(x) 2 cos(x)

    Simplify the expression:

    = -(d/dx(cos(x) sin(pi/6-x)))+d/dx(sin^2(pi/6-x))-2 cos(x) sin(x)

    Use the product rule:

    d/dx(u v) = v ( du)/( dx)+u ( dv)/( dx), where u = cos(x) and v = sin(pi/6-x):

    = d/dx(sin^2(pi/6-x))-2 cos(x) sin(x)-cos(x) d/dx(sin(pi/6-x))+d/dx(cos(x)) sin(pi/6-x)

    Simplify the expression:

    = -(cos(x) (d/dx(sin(pi/6-x))))+d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

    Using the chain rule

    d/dx(sin(pi/6-x)) = ( dsin(u))/( du) ( du)/( dx), where u = pi/6-x and ( d)/( du)(sin(u)) = cos(u):


    = d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-cos(pi/6-x) d/dx(pi/6-x) cos(x)

    Differentiate the sum term by term and factor out constants:

    = d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-d/dx(pi/6)-d/dx(x) cos(pi/6-x) cos(x)

    The derivative of pi/6 is zero:

    = d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-cos(pi/6-x) cos(x) (-(d/dx(x))+0)

    Simplify the expression:

    = cos(pi/6-x) cos(x) (d/dx(x))+d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

    The derivative of x is 1:

    = d/dx(sin^2(pi/6-x))-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+1 cos(pi/6-x) cos(x)

    Using the chain rule, d/dx(sin^2(pi/6-x)) = ( du^2)/( du) ( du)/( dx), where u = sin(pi/6-x) and ( d)/( du)(u^2) = 2 u:

    = cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+2 d/dx(sin(pi/6-x)) sin(pi/6-x)

    Using the chain rule, d/dx(sin(pi/6-x)) = ( dsin(u))/( du) ( du)/( dx), where u = pi/6-x and ( d)/( du)(sin(u)) = cos(u):

    = cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+cos(pi/6-x) d/dx(pi/6-x) 2 sin(pi/6-x)

    Differentiate the sum term by term and factor out constants:

    = cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+d/dx(pi/6)-d/dx(x) 2 cos(pi/6-x) sin(pi/6-x)

    The derivative of pi/6 again is zero:

    = cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)+2 cos(pi/6-x) sin(pi/6-x) (-(d/dx(x))+0)

    Simplify the expression:

    = cos(pi/6-x) cos(x)-2 cos(pi/6-x) (d/dx(x)) sin(pi/6-x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)

    The derivative of x is 1:

    = cos(pi/6-x) cos(x)-(d/dx(cos(x))) sin(pi/6-x)-2 cos(x) sin(x)-1 2 cos(pi/6-x) sin(pi/6-x)

    The derivative of cos(x) is -sin(x):

    = cos(pi/6-x) cos(x)-2 cos(pi/6-x) sin(pi/6-x)-2 cos(x) sin(x)--sin(x) sin(pi/6-x)

    Simplify the expression:

    Thus, the answer:

    = cos(pi/6-x) cos(x)-2 cos(pi/6-x) sin(pi/6-x)-2 cos(x) sin(x)+sin(pi/6-x) sin(x)
    Thanks from sakonpure6
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  3. #3
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    Re: Trig

    Trig-03-nov-13.png
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