Giải SBT Toán 11 KNTT Bài 33: Đạo hàm cấp 2 có đáp án

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

Có y" = (x³ – 4x)' = 3x² – 4.

Vậy y" = 3x² – 4.

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

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

$y^{\text{'}\text{'}}={\left[\frac{-3}{{\left(x-1\right)}^2}\right]}^{\text{'}}$$=-3{\left[\frac{1}{{\left(x-1\right)}^2}\right]}^{\text{'}}$$=-3\cdot \frac{-2}{{\left(x-1\right)}^3}$$=\frac{6}{{\left(x-1\right)}^3}$

Vậy $y^{\text{'}\text{'}}=\frac{6}{{\left(x-1\right)}^3}$

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

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

Vậy $y^{\text{'}\text{'}}=\frac{-4}{{\left(2x-1\right)}^2}$

b) $y^{\text{'}}={\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=\frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=\frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$1+{\tan }^2\left(x+\frac{\pi }{3}\right)$

$y^{\text{'}\text{'}}={\left[1+{\tan }^2\left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot {\left[\tan \left(x+\frac{\pi }{3}\right)\right]}^{\text{'}}$

$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{{\left(x+\frac{\pi }{3}\right)}^{\text{'}}}{co{s}^2\left(x+\frac{\pi }{3}\right)}$$=2\tan \left(x+\frac{\pi }{3}\right)\cdot \frac{1}{co{s}^2\left(x+\frac{\pi }{3}\right)}$

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

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

Có $f^{\text{'}}\left(x\right)={\left(x{e}^{x^2}+\ln \left(x+1\right)\right)}^{\text{'}}$$={\left(x{e}^{x^2}\right)}^{\text{'}}+{\left(\ln \left(x+1\right)\right)}^{\text{'}}$

$=e^{x^2}+2x^2{e}^{x^2}+\frac{1}{x+1}$

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

$=2x{e}^{x^2}+4x{e}^{x^2}+4x^3{e}^{x^2}-\frac{1}{{\left(x+1\right)}^2}$$=6x{e}^{x^2}+4x^3{e}^{x^2}-\frac{1}{{\left(x+1\right)}^2}$

Khi đó $f^{\text{'}}\left(0\right)=e^{0^2}+2\cdot 0\cdot e^{0^2}+\frac{1}{0+1}=2$

$f^{\text{'}\text{'}}\left(0\right)=6\cdot 0\cdot e^{0^2}+4\cdot 0^3{e}^{0^2}-\frac{1}{{\left(0+1\right)}^2}=-1$

Vậy f'(0) = 2 và f"(0) = −1.

Vì f(0) = 2 nên (0² + a)² + b = 2 Û a² + b = 2. (1)

Có f'(x) = 2(x² + a)×(x² + a)' = 4x×(x² + a); f"(x) = 4×(x² + a) + 8x².

Mà f"(1) = 8 nên 4×(1² + a) + 8×1² = 8 Û a + 1 = 0 Û a = −1.

Thay a = −1 vào (1), ta được (−1)² + b = 2 Û b = 1.

Vậy a = −1; b = 1.

Có $s^{\text{'}}\left(t\right)={\left[15+\sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)\right]}^{\text{'}}$$=\sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)\cdot {\left(4\pi t+\frac{\pi }{6}\right)}^{\text{'}}$

$=4\pi \sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)$

$s^{\text{'}\text{'}}\left(t\right)={\left[4\pi \sqrt{2}\cos \left(4\pi t+\frac{\pi }{6}\right)\right]}^{\text{'}}$$=-4\pi \sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)\cdot {\left(4\pi t+\frac{\pi }{6}\right)}^{\text{'}}$

$=-16{\pi }^2\sqrt{2}\sin \left(4\pi t+\frac{\pi }{6}\right)$

Gia tốc của hạt tại thời điểm t = 3 giây là

a(3) =  $s^{\text{'}\text{'}}\left(3\right)=-16{\pi }^2\sqrt{2}\sin \left(12\pi +\frac{\pi }{6}\right)$ $=-8{\pi }^2\sqrt{2}\approx -111,7$m/s².

Vậy gia tốc của hạt tại thời điểm t = 3 giây khoảng – 111,7 m/s².