Tính đạo hàm a, y = (x^2 + 2x). căn(x) b, y = (x^2 +1) / x c, y = (ax + 2b) / (3a + b) d, y = [a + (2b/x)]^3 13/09/2021 Bởi Iris Tính đạo hàm a, y = (x^2 + 2x). căn(x) b, y = (x^2 +1) / x c, y = (ax + 2b) / (3a + b) d, y = [a + (2b/x)]^3
$\displaystyle \begin{array}{{>{\displaystyle}l}} a,\ y\ =\ \left( x^{2} \ +\ 2x\right)\sqrt{x}\\ \Rightarrow y’=( 2x+2)\sqrt{x} +\frac{x^{2} +2x}{2\sqrt{x}}\\ b,\ y\ =\ \frac{x^{2} +1}{x}\\ \Rightarrow y’=\frac{2x^{2} -x^{2} +1}{x^{2}} =\frac{x^{2} +1}{x^{2}}\\ c,\ y\ =\frac{ax+2b}{3a+b}\\ \Rightarrow y’=\frac{a}{3a+b}\\ d,\ y\ =\ \left( a\ +\ \frac{2b}{x}\right)^{3}\\ \Rightarrow y’=3\left( a\ +\ \frac{2b}{x}\right)^{2}\left( -\frac{2b}{x^{2}}\right) \end{array}$ Bình luận
$\displaystyle \begin{array}{{>{\displaystyle}l}} a,\ y\ =\ \left( x^{2} \ +\ 2x\right)\sqrt{x}\\ \Rightarrow y’=( 2x+2)\sqrt{x} +\frac{x^{2} +2x}{2\sqrt{x}}\\ b,\ y\ =\ \frac{x^{2} +1}{x}\\ \Rightarrow y’=\frac{2x^{2} -x^{2} +1}{x^{2}} =\frac{x^{2} +1}{x^{2}}\\ c,\ y\ =\frac{ax+2b}{3a+b}\\ \Rightarrow y’=\frac{a}{3a+b}\\ d,\ y\ =\ \left( a\ +\ \frac{2b}{x}\right)^{3}\\ \Rightarrow y’=3\left( a\ +\ \frac{2b}{x}\right)^{2}\left( -\frac{2b}{x^{2}}\right) \end{array}$