Tìm đạo hàm của các hàm số sau y=x2 - x√x + 1

Giải Toán 11 Bài 2: Quy tắc tính đạo hàm

Bài tập 4 trang 163 SGK Toán lớp 11 Đại số:

a) $y=x^2-x\sqrt{x}+1$;

b) $y=\sqrt{2-5x-x^2}$;

c) $y=\frac{x^3}{\sqrt{a^2-x^2}}$(a là hằng số);

d) $y=\frac{1+x}{\sqrt{1-x}}$.

Lời giải:

a)

$y=x^2-x\sqrt{x}+1y^{\text{'}}={\left(x^2\right)}^{\text{'}}-{\left(x\sqrt{x}\right)}^{\text{'}}+{\left(1\right)}^{\text{'}}=2x-\left.{\left(x\right)}^{\text{'}}\sqrt{x}+x{\left(\sqrt{x}\right)}^{\text{'}}\right]+0=2x-\left(\sqrt{x}+x.\frac{1}{2\sqrt{x}}\right)=2x-\sqrt{x}-\frac{\sqrt{x}}{2}=2x-\frac{3\sqrt{x}}{2}$

b) $y=\sqrt{2-5x-x^2}$

$y^{\text{'}}=\frac{{\left(2-5x-x^2\right)}^{\text{'}}}{2\sqrt{2-5x-x^2}}=\frac{{\left(2\right)}^{\text{'}}-{\left(5x\right)}^{\text{'}}-{\left(x^2\right)}^{\text{'}}}{2\sqrt{2-5x-x^2}}=\frac{0-5-2x}{2\sqrt{2-5x-x^2}}=\frac{-2x-5}{2\sqrt{2-5x-x^2}}$

c) $y=\frac{x^3}{\sqrt{a^2-x^2}}$(a là hằng số)

$y^{\text{'}}=\frac{{\left(x^3\right)}^{\text{'}}\sqrt{a^2-x^2}-x^3{\left(\sqrt{a^2-x^2}\right)}^{\text{'}}}{{\left(\sqrt{a^2-x^2}\right)}^2}=\frac{3x^2\sqrt{a^2-x^2}-x^3\frac{{\left(a^2-x^2\right)}^{\text{'}}}{2\sqrt{a^2-x^2}}}{{\left(\sqrt{a^2-x^2}\right)}^2}=\frac{3x^2\sqrt{a^2-x^2}-x^3\frac{-2x}{2\sqrt{a^2-x^2}}}{{\left(\sqrt{a^2-x^2}\right)}^2}=\frac{3x^2\left(a^2-x^2\right)-x^4}{{\left(\sqrt{a^2-x^2}\right)}^3}=\frac{3x^2{a}^2-2x^4}{{\left(\sqrt{a^2-x^2}\right)}^3}$

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

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

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

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

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

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