Pre-interview_Test_Question_36

Screenshot 2020-07-05 at 23.23.51

Set y = x^x. Then \ln y = \ln x^x = x \ln x. Differentiating:

\displaystyle \frac 1 y \frac {\mathrm dy} {\mathrm dx} = \ln x + \frac x x = \ln x + 1

So:

\displaystyle \frac {\mathrm dy} {\mathrm dx} = y (\ln x + 1) = x^x(\ln x + 1)

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Love this one!!! :star_struck:.

Massive thanks @MathsNinja17!!!

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y = x^x = (e^{\ln x})^x = e^{x \ln x}

\dfrac{dy}{dx} = ( \ln x +1) e^{x \ln x} = ( \ln x +1) x^x

:smile:

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This is used by Imperial, amongst others, on a fairly regular basis. The first candidate of mine was asked it 20 years ago and it still crops up!

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They asked it at my oxford interview :smile:

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