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

Question

Iyana Quinn

Mathematics

12 November, 11:23

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

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Williams · 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) ].