What is the derivative of the function \(\frac{2x^2 + 3x + 7}{\sqrt{x}}\)?

[ 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)].

Answered by crisforp | 2024-06-10

The derivative of the function f ( x ) = x ​ 2 x 2 + 3 x + 7 ​ is f ′ ( x ) = 2 x 6 x 2 + 3 x − 7 ​ . This was calculated using the quotient rule and the derivatives of the numerator and denominator. Careful simplification led to the final expression for the derivative.
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Answered by crisforp | 2024-12-26