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Chain Rule, Log Differentiation & Rate of Change · Question 82 of 130

What is log y for this y = x^x?

Question 82

EasyOriginal question

Let $y = x^{x}$ for $x > 0$ .

What is $\log y$ for this $y = x^{x}$ ?

x\log x

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\log x

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x + \log x

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(\log x)^{x}

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Show the answer and explanation

CorrectNot quiteAnswer A

Correct answer: A

$\log(a^{b}) = b\log a$ .
$\log y = x\log x$ .

Worth noting

The power comes down as a factor, not as a term to add.

Concept: Taking logarithms first
Typical timeTypical: 90 s
Your timeYours: —

The rule this testsLogarithmic differentiation

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