y-x-x-y-

y-x-x-y-

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

y = x^{x}

y^{^{'}} = ?

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

y^{^{'}} = e^{x l n x} (l n x + 1)

= x^{x} (l n x + 1)

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

y^{^{'}} = x^{x} (lnx + 1)

Commented by Abdulrahman last updated on 18/May/20

thanks

Commented by Abdulrahman last updated on 18/May/20

thankx

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

\ln y = x \ln x

\frac{1}{y} y^{'} = \ln x + x \times \frac{1}{x} = 1 + \ln x)

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

Commented by Abdulrahman last updated on 18/May/20

thanks