Differentiate y = (x^x) cos2x?

I solved it and arrived at (-2x^x)(sin2x) + (x^x)cos2x(ln(x) +1). Am i even the right track? how do I know what answer to choose among the given? 

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3 Answers

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  • 2 months ago

    Use logarithmic differentiation:

    ln(y)= ln[(x^x)(cos(2x))]

    ln(y) = ln(x^x) + ln|cos(2x)|

    ln(y) = xlnx + ln|cos(2x)|

    Differentiate:

    (1/y)(dy/dx) = x(1/x) + ln(x) + (1/cos(2x))(-2sin(2x))

    (1/y)(dy/dx) = 1  + ln(x) - 2tan(2x)

    dy/dx = y[1 + ln(x) - 2tan(2x)]

    dy/dx = (x^x)(cos(2x))[1 + ln(x) - 2tan(2x)]

    dy/dx = (x^x)(cos(2x))(1 + ln(x)) - 2(x^x)sin(2x))

    Looks like the same answer you came up with.  I question the choices given.

  • Alan
    Lv 7
    2 months ago

    None of the given answers are correct. 

    Your answer appears to be correct.  

    symbolab comes up with the same answer as you. 

    https://www.symbolab.com/solver/implicit-derivativ...

    To check if your equation is equivalent to  

    other equations, just check a few values 

    like 

    0.1 

    1

    4

    However, none of the equation match your equation.   

    Something  is wrong with your textbook or course. 

  • 2 months ago

    rules:

    (x^x)' = (x^x)(1 + ln x) (cos ax)' = –a sin ax product rule (fg)' = f'g + fg' y = (x^x) cos2xf = x^xf'= (x^x)(1 + ln x) g = cos2xg' = –2sin2xy' = (x^x)(1 + ln x)(cos2x) + (x^x)(–2sin2x)y' = (x^x)(1 + ln x)(cos2x) – 2(x^x)(sin2x)y' = (x^x)(cos2x + (ln x)(cos2x)) – 2(x^x)(sin2x)y' = (x^x)(cos2x + (ln x)(cos2x) – 2sin2x)I agree with your answer, don't see how any of the given answers are correct.

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