1.

differentiate

Answer»

y= (√x)^y

take log

log y = y log√x

differentiate both side wrtx

1/y. y' = y'log√x + 1/2√x. 1/√x .y

1/y. y'= y' log√x + 1/2x . y

y'( 1/y - log √x) = y/2x

y'= y/2x( 1/y - log √x)



Discussion

No Comment Found