Question

What is the derivative of the function (2x^2+3x+7)/(sqrt(x))

Answer 1

[ 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)].