Tính các đạo hàm sau: a. `y=x^3/3-x^2/2+2x-5` b. `y=2x^4+3x-2/x` c. `y=mx^3+(4+n)x^2+8“\sqrt[x]+2\sqrt[2x-1]+n` (m,n là hằng số) d. `y=1-6\sqrt[x]+

Đáp án:

$\begin{array}{l}
a)y = \frac{{{x^3}}}{3} – \frac{{{x^2}}}{2} + 2x – 5\\
 \Rightarrow y’ = {x^2} – x + 2\\
b)y = 2{x^4} + 3x – \frac{2}{x}\\
 \Rightarrow y’ = 8{x^3} + 3 + \frac{2}{{{x^2}}}\\
c)y = m{x^3} + \left( {4 + n} \right){x^2} + 8\sqrt x  + 2\sqrt {2x – 1}  + n\\
 \Rightarrow y’ = 3m{x^2} + 2\left( {4 + n} \right)x + \frac{4}{{\sqrt x }} + \frac{2}{{\sqrt {2x – 1} }}\\
d)y = 1 – 6\sqrt x  + \frac{2}{{x + 1}} – \frac{3}{x} + 3\sqrt {2x + 1} \\
 \Rightarrow y’ =  – \frac{3}{{\sqrt x }} – \frac{2}{{{{\left( {x + 1} \right)}^2}}} + \frac{3}{{{x^2}}} + \frac{3}{{\sqrt {2x + 1} }}\\
e)y = \frac{{{x^3}}}{3} – x\sqrt x  + 3x + 1 + \frac{2}{{\sqrt x }} + \frac{5}{x}\\
 \Rightarrow y’ = {x^2} – \frac{3}{2}\sqrt x  + 3 – \frac{1}{{x\sqrt x }} – \frac{5}{{{x^2}}}
\end{array}$