Tính đạo hàm của các hàm số sau: a) y=-3x^2/2+2/x+x^3/3

Giải SBT Toán 11 Bài 2: Các quy tắc tính đạo hàm

Bài 1 trang 43 SBT Toán 11 Tập 2: Tính đạo hàm của các hàm số sau:

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

b) y = (x² − 1)(x² – 4)(x² + 9);

c) $y=\frac{x^2-2x}{x^2+x+1}$;

d) $y=\frac{1-2x}{x+1}$;

e) y = xe^{2x + 1};

g) y = (2x + 3)3^{2x + 1};

h) y = xln²x;

i) $y={\log }_2\left(x^2+1\right)$.

Lời giải:

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

b) y = (x² − 1)(x² – 4)(x² + 9)

= (x⁴ – 5x² + 4)(x² + 9)

= x⁶ – 5x⁴ + 4x² + 9x⁴ – 45x² + 36

= x⁶ + 4x⁴ – 41x² + 36.

y' = 6x⁵ + 16x³ – 82x

c) $y^{\text{'}}=\frac{{\left(x^2-2x\right)}^{\text{'}}\left(x^2+x+1\right)-\left(x^2-2x\right){\left(x^2+x+1\right)}^{\text{'}}}{{\left(x^2+x+1\right)}^2}$

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

$=\frac{\left(2x^3-2\right)-\left(2x^3-3x^2-2x\right)}{{\left(x^2+x+1\right)}^2}=\frac{3x^2+2x-x}{{\left(x^2+x+1\right)}^2}$.

d) $y^{\text{'}}=\frac{{\left(1-2x\right)}^{\text{'}}\left(x+1\right)-\left(1-2x\right){\left(x+1\right)}^{\text{'}}}{{\left(x+1\right)}^2}$

$=\frac{\left(-2\right)\left(x+1\right)-\left(1-2x\right)}{{\left(x+1\right)}^2}=-\frac{3}{{\left(x+1\right)}^2}.$

e) $y\text{'}=x\text{'}{e}^{2x+1}+x\left(e^{2x+1}\right)\text{'}=e^{2x+1}+x.{\left(2x+1\right)}^{\text{'}}.e^{2x+1}$

$=e^{2x+1}+x\mathrm{.2.}{e}^{2x+1}=\left(2x+1\right)e^{2x+1}$.

g) $y^{\text{'}}=\left(2x+3\right)3^{2x+1}={\left(2x+3\right)}^{\text{'}}{3}^{2x+1}+\left(2x+3\right){\left(3^{2x+1}\right)}^{\text{'}}$

$={2.3}^{2x+1}+\left(2x+3\right){\left(2x+1\right)}^{\text{'}}{3}^{2x+1}.\ln 3$

$={2.3}^{2x+1}+\left(2x+3\right){\mathrm{.2.3}}^{2x+1}.\ln 3$

h) $y^{\text{'}}=x^{\text{'}}l{n}^{2}x+x{\left(l{n}^{2}x\right)}^{\text{'}}={\ln }^{2}x+x.2\ln x.\frac{1}{x}$

$={\ln }^{2}x+2\ln x$.

i) $y\text{'}={\log }_2\left(x^2+1\right)=\frac{{\left(x^2+1\right)}^{\text{'}}}{\left(x^2+1\right)\ln 2}=\frac{{\left(x^2+1\right)}^{\text{'}}}{\left(x^2+1\right)\ln 2}=\frac{2x}{\left(x^2+1\right)\ln 2}$.