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

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

Có y*"* = (x$^{3}$ – 4x)' = 3x$^{2}$ – 4.

Vậy y*"* = 3x$^{2}$ – 4.

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

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

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

Vậy ${y}^{''}=\frac{6}{{(x-1)}^{3}}$

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

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

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

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

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

$=2\tan (x+\frac{π}{3})\cdot \frac{{(x+\frac{π}{3})}^{'}}{co{s}^{2}(x+\frac{π}{3})}$$=2\tan (x+\frac{π}{3})\cdot \frac{1}{co{s}^{2}(x+\frac{π}{3})}$

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

Vậy ${y}^{''}=2\tan (x+\frac{π}{3})+2{\tan }^{3}(x+\frac{π}{3})$

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

$={e}^{{x}^{2}}+x{e}^{{x}^{2}}\cdot {({x}^{2})}^{'}+\frac{1}{x+1}$

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

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

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

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

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

Vậy f*'(0) = 2 và f"*(0) = −1.

Vì f(0) = 2 nên (0$^{2}$ + a)$^{2}$ + b = 2 Û a$^{2}$ + b = 2. (1)

Có f*'(x) = 2(x$^{2}$ + a)×(x$^{2}$ + a)'* = 4x×(x$^{2}$ + a); f*"*(x) = 4×(x$^{2}$ + a) + 8x$^{2}$.

Mà f*"*(1) = 8 nên 4×(1$^{2}$ + a) + 8×1$^{2}$ = 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}^{'}(t)={[15+\sqrt{2}\sin (4πt+\frac{π}{6})]}^{'}$$=\sqrt{2}\cos (4πt+\frac{π}{6})\cdot {(4πt+\frac{π}{6})}^{'}$

$=4π\sqrt{2}\cos (4πt+\frac{π}{6})$

${s}^{''}(t)={[4π\sqrt{2}\cos (4πt+\frac{π}{6})]}^{'}$$=-4π\sqrt{2}\sin (4πt+\frac{π}{6})\cdot {(4πt+\frac{π}{6})}^{'}$

$=-16{π}^{2}\sqrt{2}\sin (4πt+\frac{π}{6})$

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

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

Vậy gia tốc của hạt tại thời điểm t = 3 giây khoảng – 111,7 m/s$^{2}$.