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What is the derivative of y = x^(lnx)? Show your solution.

User Nyxm
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2 Answers

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y= x^(\ln x) = \left ( e^(\ln x) \right )^(\ln x) = e^(\ln x \cdot \ln x) = e^{\ln^(2) x} \\ \\ y' = \left ( e^{\ln^(2) x} \right )' = e^{\ln^(2) x} \cdot \left ( \ln^(2) x \right )' = e^{\ln^(2) x} \cdot 2 \ln x \cdot (\ln x)' = \\ \\ = \frac{ e^{\ln^(2) x} \cdot 2 \ln x }{x} = \boxed{ (2\cdot x^(\ln x) \cdot \ln x )/(x) }
User Sondes
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4 votes
Hello,

ln(y)=ln(x^(ln(x))=ln(x)*ln(x)=(ln(x))²

(ln(y))'=2ln(x)*1/x
(1/y)*y'=2ln(x) / x
y'=(2 ln(x) * x^(ln(x)) ) /x
User Phym
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