Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

Question Number 94424 by Abdulrahman last updated on 18/May/20

y=x^x   y^′ =?

$$\mathrm{y}=\mathrm{x}^{\mathrm{x}} \\ $$$$\mathrm{y}^{'} =? \\ $$

Commented by Tony Lin last updated on 18/May/20

y^′ =e^(xlnx) (lnx+1)     =x^x (lnx+1)

$${y}^{'} ={e}^{{xlnx}} \left({lnx}+\mathrm{1}\right) \\ $$$$\:\:\:={x}^{{x}} \left({lnx}+\mathrm{1}\right) \\ $$

Commented by PRITHWISH SEN 2 last updated on 18/May/20

y^′ =x^x (lnx+1)

$$\mathrm{y}^{'} =\mathrm{x}^{\mathrm{x}} \left(\mathrm{lnx}+\mathrm{1}\right) \\ $$

Commented by Abdulrahman last updated on 18/May/20

thanks

$$\mathrm{thanks} \\ $$

Commented by Abdulrahman last updated on 18/May/20

thankx

$$\mathrm{thankx} \\ $$

Answered by prakash jain last updated on 18/May/20

ln y=xln x  (1/y)y′=ln x+x×(1/x)=1+ln x)  y′=y(1+ln x)=x^x (1+ln x)

$$\mathrm{ln}\:{y}={x}\mathrm{ln}\:{x} \\ $$$$\left.\frac{\mathrm{1}}{{y}}{y}'=\mathrm{ln}\:{x}+{x}×\frac{\mathrm{1}}{{x}}=\mathrm{1}+\mathrm{ln}\:{x}\right) \\ $$$${y}'={y}\left(\mathrm{1}+\mathrm{ln}\:{x}\right)={x}^{{x}} \left(\mathrm{1}+\mathrm{ln}\:{x}\right) \\ $$

Commented by Abdulrahman last updated on 18/May/20

thanks

$$\mathrm{thanks} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com