Question

d/dx(xlog ex-x) iss ka answer please fast​

Answer 1

I am assuming it is d/dx [ x*log( $e^{x}$-x ) ]

In that case first we use (UV)' and then apply chain rule.

So, U is x , V is log( $e^{x}$-x ).

Hence, U'V + V'U

log( $e^{x}$-x ) + [x/( $e^{x}$-x )] ( $e^{x}$ - 1) -------------- Applying chain rule.

So, the final answer is, log( $e^{x}$-x ) +[ x( $e^{x}$ - 1) ]/ ( $e^{x}$-x )

Taking LCM,

The final answer is {log${{(e^{x} - x)}^{e^{x} -x}$ + [ x( $e^{x}$ - 1) ]} / ( $e^{x}$-x )