Tìm đạo hàm của các hàm số sau Bài tập 1 trang 176 SGK Toán lớp 11 Đại số

Giải Toán 11 Ôn tập chương 5

Bài tập 1 trang 176 SGK Toán lớp 11 Đại số: Tìm đạo hàm của các hàm số sau:

a) $y=\frac{x^3}{3}-\frac{x^2}{2}+x-5$;

b) $y=\frac{2}{x}-\frac{4}{x^2}+\frac{5}{x^3}-\frac{6}{7x^4}$;

c) $y=\frac{3x^2-6x+7}{4x}$;

d) $y=\left(\frac{2}{x}+3x\right)\left(\sqrt{x}-1\right)$;

e) $y=\frac{1+\sqrt{x}}{1-\sqrt{x}}$;

f) $y=\frac{-x^2+7x+5}{x^2-3x}$.

Lời giải:

a) $y=\frac{x^3}{3}-\frac{x^2}{2}+x-5$

$y^{\text{'}}={\left(\frac{x^3}{3}\right)}^{\text{'}}-{\left(\frac{x^2}{2}\right)}^{\text{'}}+(x{)}^{\text{'}}-(5{)}^{\text{'}}=\frac{3x^2}{3}-\frac{2x}{2}+1$= x² – x + 1.

b) $y=\frac{2}{x}-\frac{4}{x^2}+\frac{5}{x^3}-\frac{6}{7x^4}$

$\begin{array}{l}{y}^{\text{'}}={\left(\frac{2}{x}\right)}^{\text{'}}-{\left(\frac{4}{x^2}\right)}^{\text{'}}+{\left(\frac{5}{x^3}\right)}^{\text{'}}-{\left(\frac{6}{7x^4}\right)}^{\text{'}}\\ =-\frac{2}{x^2}-\frac{-4\cdot {\left(x^2\right)}^{\text{'}}}{x^4}+\frac{-5{\left(x^3\right)}^{\text{'}}}{x^6}-\frac{-6{\left(x^4\right)}^{\text{'}}}{7x^8}\\ =-\frac{2}{x^2}+\frac{4.2x}{x^4}-\frac{5.3x^2}{x^6}+\frac{6.4x^3}{7x^8}\\ =-\frac{2}{x^2}+\frac{8}{x^3}-\frac{15}{x^4}+\frac{24}{7x^5}\end{array}$

c)$y=\frac{3x^2-6x+7}{4x}$

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(3x^2-6x+7\right)}^{\text{'}}.4x-\left(3x^2-6x+7\right).(4x{)}^{\text{'}}}{{\left(4x\right)}^2}\\ =\frac{(6x-6).4x-4\left(3x^2-6x+7\right)}{16x^2}\\ =\frac{24x^2-24x-12x^2+24x-28}{16x^2}\\ =\frac{12x^2-28}{16x^2}=\frac{3x^2-7}{4x^2}\end{array}$

Cách khác:

$\begin{array}{l}y=\frac{3}{4}x-\frac{3}{2}+\frac{7}{4x}\\ y^{\text{'}}={\left(\frac{3}{4}x\right)}^{\text{'}}-{\left(\frac{3}{2}\right)}^{\text{'}}+{\left(\frac{7}{4x}\right)}^{\text{'}}=\frac{3}{4}-0-\frac{7}{4x^2}=\frac{3x^2-7}{4x^2}\end{array}$

d) $y=\left(\frac{2}{x}+3x\right)\left(\sqrt{x}-1\right)$

$\begin{array}{l}{y}^{\text{'}}={\left(\frac{2}{x}+3x\right)}^{\text{'}}\left(\sqrt{x}-1\right)+\left(\frac{2}{x}+3x\right){\left(\sqrt{x}-1\right)}^{\text{'}}\\ =\left(-\frac{2}{x^2}+3\right)\left(\sqrt{x}-1\right)+\left(\frac{2}{x}+3x\right)\cdot \frac{1}{2\sqrt{x}}\\ =\frac{-2}{x\sqrt{x}}+\frac{2}{x^2}+3\sqrt{x}-3+\frac{1}{x\sqrt{x}}+\frac{3}{2}\sqrt{x}\\ =\frac{-1}{x\sqrt{x}}+\frac{2}{x^2}+\frac{9\sqrt{x}}{2}-3\end{array}$

e) $y=\frac{1+\sqrt{x}}{1-\sqrt{x}}$

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(1+\sqrt{x}\right)}^{\text{'}}\left(1-\sqrt{x}\right)-\left(1+\sqrt{x}\right){\left(1-\sqrt{x}\right)}^{\text{'}}}{{\left(1-\sqrt{x}\right)}^2}\\ =\frac{\frac{1}{2\sqrt{x}}\left(1-\sqrt{x}\right)+\frac{1}{2\sqrt{x}}\left(1+\sqrt{x}\right)}{{\left(1-\sqrt{x}\right)}^2}\\ =\frac{1}{\sqrt{x}{\left(1-\sqrt{x}\right)}^2}\end{array}$

f) $y=\frac{-x^2+7x+5}{x^2-3x}$

$\begin{array}{l}{y}^{\text{'}}=\frac{{\left(-x^2+7x+5\right)}^{\text{'}}\left(x^2-3x\right)-\left(-x^2+7x+5\right){\left(x^2-3x\right)}^{\text{'}}}{{\left(x^2-3x\right)}^2}\\ =\frac{(-2x+7)\left(x^2-3x\right)-(2x-3)\left(-x^2+7x+5\right)}{{\left(x^2-3x\right)}^2}\\ =\frac{-2x^3+13x^2-21x+2x^3-17x^2+11x+15}{{\left(x^2-3x\right)}^2}\\ =\frac{-4x^2-10x+15}{{\left(x^2-3x\right)}^2}\end{array}$