Verified Answer :

[ sqrt(x)] ' = 1 / [2sqrt(x)] ;
( x^2 )' = 2x ;
2' = 0 ;
x ' = 1 ;
3 ' = 0 ;
7 ' = 0 ;
( f g ) ' = f ' g + f g '  and ( f + g ) ' = f ' + g ' => (2x^2+3x+7) ' = 4x + 3;but, ( f / g ) ' = [  f ' g - f g ' ] /  (g^2) => [(2x^2+3x+7)/(sqrt(x))] ' = {( 4x + 3 ) sqrt(x) - (2x^2+3x+7)   1 / [2sqrt(x)] } / x = [ (2x)(4x + 3 ) -  (2x^2+3x+7) ] / [ x  2sqrt(x) ] = ( 6x^2 + 3x - 7 ) / [ 2 x sqrt(x)].