Dùng định nghĩa để tính đạo hàm của các hàm số sau: f(x)=căn (4x+1) tại x = 2

Giải SBT Toán 11 Bài tập cuối chương 7

Bài 1 trang 45 SBT Toán 11 Tập 2: Dùng định nghĩa để tính đạo hàm của các hàm số sau:

a) $f\left(x\right)=\sqrt{4x+1}$ tại x = 2;

b) $f\left(x\right)=x^4$ tại x = ‒1;

c) $f\left(x\right)=\frac{1}{x+1}$;

d) $f\left(x\right)=\sqrt[3]{x^2+1}$

Lời giải:

a) Với $x_0\ge -\frac{1}{4}$, ta có:

$y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\sqrt{4x+1}-\sqrt{4x_0+1}}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{4\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)}{\left(4x+1\right)-\left(4x_0+1\right)}$

$=\underset{x\to x_0}{\lim }\frac{4\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)}{\left(\sqrt{4x+1}-\sqrt{4x_0+1}\right)\left(\sqrt{4x+1}+\sqrt{4x_0+1}\right)}$

$=\underset{x\to x_0}{\lim }\frac{4}{\left(\sqrt{4x+1}+\sqrt{4x_0+1}\right)}$

$=\frac{4}{2\sqrt{4x_0+1}}$.

Vậy $y^{\text{'}}\left(2\right)=\frac{4}{2\sqrt{4.2+1}}=\frac{2}{3}$.

b) Với $x_0\in$ ℝ, ta có:

$y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{x^4-{x_0}^4}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{x^4-{x_0}^4}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x^2-{x_0}^2\right)\left(x^2+{x_0}^2\right)}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{\left(x-x_0\right)\left(x+x_0\right)\left(x^2+{x_0}^2\right)}{x-x_0}$

$=\underset{x\to x_0}{\lim }\left(x+x_0\right)\left(x^2+{x_0}^2\right)=2x_0.2{x_0}^2=4{x_0}^3$

Vậy $y^{\text{'}}(-1)=4.{(-1)}^3=-4$.

c) Với $x_0\ne -1$, ta có:

$y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{\frac{1}{x+1}-\frac{1}{x_0+1}}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\left(x_0+1\right)-\left(x+1\right)}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}$

$=\underset{x\to x_0}{\lim }\frac{x_0-x}{\left(x-x_0\right)\left(x+1\right)\left(x_0+1\right)}=\underset{x\to x_0}{\lim }-\frac{1}{\left(x+1\right)\left(x_0+1\right)}$

$=-\frac{1}{{\left(x_0+1\right)}^2}$.

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

d) Với $x_0\in$ ℝ, ta có:

$y^{\text{'}}\left(x_0\right)=\underset{x\to x_0}{\lim }\frac{f\left(x\right)-f\left(x_0\right)}{x-x_0}=\underset{x\to x_0}{\lim }\frac{\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}}{x-x_0}$

$=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(x-x_0\right)\left(x+x_0\right)}$

$=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{x^2-x_0^2}$$=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(x^2+1\right)-\left(x_0^2+1\right)}$

$=\underset{x\to x_0}{\lim }\frac{\left(x+x_0\right)\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)}{\left(\sqrt[3]{x^2+1}-\sqrt[3]{x_0^2+1}\right)\left(\sqrt[3]{{\left(x^2+1\right)}^2}+\sqrt[3]{x^2+1}\sqrt[3]{x_0^2+1}+\sqrt[3]{{\left(x_0^2+1\right)}^2}\right)}$

$=\underset{x\to x_0}{\lim }\frac{x+x_0}{\sqrt[3]{{\left(x^2+1\right)}^2}+\sqrt[3]{x^2+1}\sqrt[3]{x_0^2+1}+\sqrt[3]{{\left(x_0^2+1\right)}^2}}$$=\underset{x\to x_0}{\lim }\frac{2x_0}{3\sqrt[3]{{\left(x_0^2+1\right)}^2}}$.

Vậy $y^{\text{'}}\left(x\right)=\frac{2x}{3\sqrt[3]{{\left(x^2+1\right)}^2}}$ ($x\in$ ℝ).