Cách làm cho bạn:
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 \(dy=[(2x + 4)(x^2- \sqrt x) + (x^2+ 4x + 1)(2x - \frac{1}{2\sqrt{x}})]dx\)
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