Derivative

[AvgStudent]

How do I differentiate [imath]x^x?[/imath]

 

[Deleted member 4993]

How do I differentiate [imath]x^x?[/imath]

y = x^x........Now Use 'log'

log(y) = x * log(x) ...... Now differentiate using chain-rule and product rule.

 

[Dr.Peterson]

How do I differentiate [imath]x^x?[/imath]

Alternatively, you can write [imath]x^x=\left(e^{\ln(x)}\right)^x=e^{x\ln(x)}[/imath] and use the chain rule.

 

[AvgStudent]

y = x^x........Now Use 'log'

log(y) = x * log(x) ...... Now differentiate using chain-rule and product rule.

Is this right? I get the same answer both ways.
[math]y=x^x\\ \log(y)=x\log(x)\\ \dfrac{d}{dx}\log(y)=\dfrac{d}{dx}x\log(x)\\ \dfrac{1}{y}\cdot y'=\log(x)+x\cdot \dfrac{1}{x}\\ y'=y\cdot(\log(x)+1)=x^x(\log(x)+1)[/math]

Alternatively, you can write [imath]x^x=\left(e^{\ln(x)}\right)^x=e^{x\ln(x)}[/imath] and use the chain rule.

[math]x^x=\left(e^{\ln(x)}\right)^x=e^{x\ln(x)}\\ \frac{d}{dx}e^{x\ln(x)}=e^{x\ln(x)}(\ln(x)+x\cdot \frac{1}{x})=x^x(\ln(x)+1)[/math]

 

[blamocur]

Is this right? I get the same answer both ways.
[math]y=x^x\\ \log(y)=x\log(x)\\ \dfrac{d}{dx}\log(y)=\dfrac{d}{dx}x\log(x)\\ \dfrac{1}{y}\cdot y'=\log(x)+x\cdot \dfrac{1}{x}\\ y'=y\cdot(\log(x)+1)=x^x(\log(x)+1)[/math]
[math]x^x=\left(e^{\ln(x)}\right)^x=e^{x\ln(x)}\\ \frac{d}{dx}e^{x\ln(x)}=e^{x\ln(x)}(\ln(x)+x\cdot \frac{1}{x})=x^x(\ln(x)+1)[/math]

Just being fussy: [imath]\ln[/imath] and [imath]\log[/imath] are the same function in the above.

 

[Steven G]

Yes, your work is fine assuming that by log you mean natural log or log_e