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) \(y =  \frac{1+x}{\sqrt{1-x}}\)

\(y' =  \frac{\left ( 1+x \right )'.\sqrt{1-x}-\left ( 1+x \right ).\left ( \sqrt{1-x} \right )'}{1-x}\)

\(= \frac{\sqrt{1-x}-\left ( 1+x \right )\frac{-1}{2\sqrt{1-x}}}{1-x}\)

\( =\frac{2\left ( 1-x \right )+1+x}{2\left ( 1-x \right )\sqrt{1-x}}\)

\( =\frac{3-x}{2\left ( 1-x \right )\sqrt{1-x}}\)