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Derivative of d/dx(x^x)?

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Derivative of d/dx(x^x)?
asked Jun 28, 2013 in CALCULUS by payton Apprentice

1 Answer

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d/dx(x^x) = d/dx(e^log(x^x))                 [Expressing  x^x as a power of e, we get  x^x = e^log(x^x)]

                = d/dx(e^(xlogx))

                = e^(xlogx) [ d/dx(xlogx) ]   [Applying chain rule]

                = x^x [ d/dx(xlogx) ]      [Epressing e^(xlogx) in terms of x, we get e^(xlogx)=e^log(x^x)= x^x]

                = x^x [ logx d/dx(x) + x d/dx(logx) ]     [Since d/dx uv = vdu/dx + udv/dx]

                = x^x [ logx(1) + x(1/x) ]      [Since d/dx(x) = 1 and d/dx(logx) = 1/x]

                = x^x (logx +1)

Therefore d/dx(x^x) = x^x (logx +1)

 

answered Jun 28, 2013 by joly Scholar

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