If f(x) = x^x + (log x)^x for x > 1, what is f'(e)?

Question 102

HardOriginal question

If $f(x) = x^{x} + (\log x)^{x}$ for $x > 1$ , what is $f'\left(e\right)$ ?

e^{e} + 1

Correct answerYour answer

2e^{e}

Correct answerYour answer

2e^{e} + 1

Correct answerYour answer

3e^{e}

Correct answerYour answer

Show the answer and explanation

CorrectNot quiteAnswer C

Correct answer: C

Differentiate each term on its own: $\frac{d}{dx}x^{x} = x^{x}(1 + \log x)$ ; for $u = (\log x)^{x}$ , $\log u = x\log(\log x)$ gives $u' = (\log x)^{x}\left(\log(\log x) + \frac{1}{\log x}\right)$ .
At $x = e$ : $e^{e}(1 + 1) + 1 \times (0 + 1) = 2e^{e} + 1$ .

Worth noting

$3e^{e}$ adds the logarithms of the two terms, as if $f$ were their product; a sum is differentiated term by term.

Ten questions from Chain Rule, Log Differentiation & Rate of Change, at your own pace.