Giải câu 1 bài 4: Vi phân

a) \(y = \frac{\sqrt{x}}{a+b}\) (\(a, b\) là hằng số)

\(dy = d \left ( \frac{\sqrt{x}}{a+b} \right ) =  \left ( \frac{\sqrt{x}}{a+b} \right )dx = \frac{1}{2(a+b)\sqrt{x}}dx\).

Vậy \(dy =\frac{1}{2(a+b)\sqrt{x}}dx\)

b) \(y = (x^2+ 4x + 1)(x^2- \sqrt x)\)

\(dy = d(x^2+ 4x + 1)(x^2- \sqrt x) \)

\(= [(2x + 4)(x^2- \sqrt x) + (x^2+ 4x + 1)(2x -  \frac{1}{2\sqrt{x}})]dx\).

Vậy \(dy=[(2x + 4)(x^2- \sqrt x) + (x^2+ 4x + 1)(2x -  \frac{1}{2\sqrt{x}})]dx\)