Giải SBT Toán 11 KNTT Bài 32: Các quy tắc tính đạo hàm có đáp án

a) Ta có: y' = ((x + 1)²)'(x² – 1) + (x + 1)²(x² – 1)'

= 2(x + 1)(x² – 1) + 2x(x + 1)²

= 2x³ – 2x + 2x² – 2 + 2x³ + 4x² + 2x = 4x³ + 6x² – 2.

Vậy y' = 4x³ + 6x² – 2.

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

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

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

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

$=\frac{\left(2x-1\right)\cdot \left(x+2\right)-\left(x^2-x+1\right)}{{\left(x+2\right)}^2}$$=\frac{2x^2+3x-2-x^2+x-1}{{\left(x+2\right)}^2}$$=\frac{x^2+4x-3}{{\left(x+2\right)}^2}$

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

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

 $=\frac{{\left(1-x^2\right)}^{\text{'}}\cdot \left(x^2+1\right)-\left(1-x^2\right)\cdot {\left(x^2+1\right)}^{\text{'}}}{{\left(x^2+1\right)}^2}$$=\frac{\left(-2x\right)\cdot \left(x^2+1\right)-\left(1-x^2\right)\cdot \left(2x\right)}{{\left(x^2+1\right)}^2}$$=\frac{-2x^3-2x-\left(2x-2x^3\right)}{{\left(x^2+1\right)}^2}$$=\frac{-4x}{{\left(x^2+1\right)}^2}$

Có $f^{\text{'}}\left(x\right)={\left(\frac{x}{\sqrt{4-x^2}}\right)}^{\text{'}}$ $=\frac{x^{\text{'}}\cdot \sqrt{4-x^2}-x\cdot {\left(\sqrt{4-x^2}\right)}^{\text{'}}}{4-x^2}$

$=\frac{\sqrt{4-x^2}-x\cdot \frac{-2x}{2\sqrt{4-x^2}}}{4-x^2}$$=\frac{\sqrt{4-x^2}+\frac{x^2}{\sqrt{4-x^2}}}{4-x^2}$

$=\frac{4-x^2+x^2}{\left(4-x^2\right)\sqrt{4-x^2}}$$=\frac{4}{\left(4-x^2\right)\sqrt{4-x^2}}$

Khi đó $f^{\text{'}}\left(0\right)=\frac{4}{\left(4-0\right)\sqrt{4-0}}=\frac{1}{2}$

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

Khi đó $g^{\text{'}}\left(1\right)=-\frac{1}{1^2}-\frac{1}{2.1\sqrt{1}}+2\cdot 1=\frac{1}{2}$

Do đó f'(0) – g'(1) = $\frac{1}{2}-\frac{1}{2}=0$ . Vậy f'(0) – g'(1) = 0.

Có $y^{\text{'}}={\left(3\tan \left(x+\frac{\pi }{4}\right)-2\cot \left(\frac{\pi }{4}-x\right)\right)}^{\text{'}}$

$={\left(3\tan \left(x+\frac{\pi }{4}\right)\right)}^{\text{'}}-{\left(2\cot \left(\frac{\pi }{4}-x\right)\right)}^{\text{'}}$

$=3\cdot \frac{{\left(x+\frac{\pi }{4}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{4}\right)}+2\cdot \frac{{\left(\frac{\pi }{4}-x\right)}^{\text{'}}}{{\sin }^2\left(\frac{\pi }{4}-x\right)}$

$=\frac{3}{co{s}^2\left(x+\frac{\pi }{4}\right)}-\frac{2}{{\sin }^2\left(\frac{\pi }{4}-x\right)}$

$=\frac{3}{co{s}^2\left(x+\frac{\pi }{4}\right)}-\frac{2}{{\sin }^2\left[\frac{\pi }{2}-\left(\frac{\pi }{4}+x\right)\right]}$

$=\frac{3}{co{s}^2\left(x+\frac{\pi }{4}\right)}-\frac{2}{co{s}^2\left(x+\frac{\pi }{4}\right)}$

 $=\frac{1}{co{s}^2\left(x+\frac{\pi }{4}\right)}$$=1+{\tan }^2\left(x+\frac{\pi }{4}\right)$

+ Có $f^{\text{'}}\left(x\right)={\left[4{\sin }^2\left(2x-\frac{\pi }{3}\right)\right]}^{\text{'}}$$=8\sin \left(2x-\frac{\pi }{3}\right){\left[\sin \left(2x-\frac{\pi }{3}\right)\right]}^{\text{'}}$

$=8\sin \left(2x-\frac{\pi }{3}\right)cos\left(2x-\frac{\pi }{3}\right)\cdot {\left(2x-\frac{\pi }{3}\right)}^{\text{'}}$

$=8\cdot 2\sin \left(2x-\frac{\pi }{3}\right)cos\left(2x-\frac{\pi }{3}\right)$$=8\sin \left(4x-\frac{2\pi }{3}\right)$

Vì $\left|\sin \left(4x-\frac{2\pi }{3}\right)\right|\le 1$  với mọi x Î ℝ nên $\left|8\sin \left(4x-\frac{2\pi }{3}\right)\right|\le 8$  với mọi x Î ℝ .

Vậy |f'(x)| ≤ 8 với mọi x Î ℝ.

+ Có f'(x) = 8 $\iff 8\sin \left(4x-\frac{2\pi }{3}\right)=8$

 $\iff \sin \left(4x-\frac{2\pi }{3}\right)=1$

 $\iff 4x-\frac{2\pi }{3}=\frac{\pi }{2}+k2\pi$ (k Î ℤ)

  $\iff 4x=\frac{7\pi }{6}+k2\pi$ (k Î ℤ)

$\iff x=\frac{7\pi }{24}+k\frac{\pi }{2}$ (k Î ℤ).

Vậy f'(x) = 8 khi $x=\frac{7\pi }{24}+k\frac{\pi }{2}$ với k Î ℤ.