Giải câu 4 bài 2: Quy tắc tính đạo hàm.
a) \(y = x^2 - x\sqrt x + 1\)
\(y' = 2x-[x'\sqrt{x}+x(\sqrt{x})']=2x - \left ( \sqrt{x}+x.\frac{1}{2\sqrt{x}} \right )= 2x - \frac{3}{2}\sqrt{x}\)
b) \(y = \sqrt {(2 - 5x - x^2)}\)
\(y' =\frac{\left ( 2-5x-x^{2} \right )'}{2.\sqrt{2-5x-x^{2}}}= \frac{-5-2x}{2\sqrt{2-5x-x^{2}}}\)
c) \(y = \frac{x^{3}}{\sqrt{a^{2}-x^{2}}}\) ( \(a\) là hằng số)
\(y' = \frac{( x^{3})'.\sqrt{a^{2}-x^{2}}-x^{3}.\left ( \sqrt{a^{2}-x^{2}} \right )}{a^{2}-x^{2}}\)
\(= \frac{3x^{2}.\sqrt{a^{2}-x^{2}}-x^{3}.\frac{-2x}{2\sqrt{a^{2}-x^{2}}}}{a^{2}-x^{2}}\)
\( =\frac{3x^{2}.\sqrt{a^{2}-x^{2}}+\frac{x^{4}}{\sqrt{a^{2}-x^{2}}}}{a^{2}-x^{2}}\)
\(= \frac{x^{2}\left ( 3a^{2}-2x^{2} \right )}{\left ( a^{2} -x^{2}\right )\sqrt{a^{2}-x^{2}}}\)
d) \(y = \frac{1+x}{\sqrt{1-x}}\)
\(y' = \frac{\left ( 1+x \right )'.\sqrt{1-x}-\left ( 1+x \right ).\left ( \sqrt{1-x} \right )'}{1-x}\)
\(= \frac{\sqrt{1-x}-\left ( 1+x \right )\frac{-1}{2\sqrt{1-x}}}{1-x}\)
\( =\frac{2\left ( 1-x \right )+1+x}{2\left ( 1-x \right )\sqrt{1-x}}\)
\( =\frac{3-x}{2\left ( 1-x \right )\sqrt{1-x}}\)