Expert Answer to "Differentiate, please: y=xx."

Julia White

Answered question

2021-12-09

Differentiate, please:

y = x^{x}

Virginia Palmer

Beginner2021-12-10Added 27 answers

Use logarithmic diferentiation:

\ln y = {\ln x}^{x}

Diferentiaate both sides with respect to x:

\frac{1}{y} \frac{d y}{d x} = \ln x + x \frac{1}{x}

\frac{1}{y} \frac{d y}{d x} = \ln x + 1

Multiply by y:

\frac{d y}{d x} = y (\ln x + 1)

Now

y = x^{x}

, thus,

\frac{d y}{d x} = x^{x} (\ln x + 1)

Jeremy Merritt

Beginner2021-12-11Added 31 answers

Apply exponent rule:

a^{b} = e^{b \ln (a)}

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

= {(\frac{d}{d x})}^{e^{x \ln (x)}}

Now, apply the Chain Rule:

= {(\frac{d}{d x})}^{e^{x \ln (x)}} = e^{x \ln (x)} \frac{d}{d x} (x \ln (x))

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

x^{x} (\ln (x) + 1)

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