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

a) Ta có: y*'* = ((x + 1)$^{2}$)'(x$^{2}$ – 1) + (x + 1)$^{2}$(x$^{2}$ – 1)'

= 2(x + 1)(x$^{2}$ – 1) + 2x(x + 1)$^{2}$

= 2x$^{3}$ – 2x + 2x$^{2}$ – 2 + 2x$^{3}$ + 4x$^{2}$ + 2x = 4x$^{3}$ + 6x$^{2}$ – 2.

Vậy y' = 4x$^{3}$ + 6x$^{2}$ – 2.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$={(3\tan (x+\frac{π}{4}))}^{'}-{(2\cot (\frac{π}{4}-x))}^{'}$

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

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

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

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

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

Có ${f}^{'}(x)={(co{s}^{2}x+co{s}^{2}(\frac{2π}{3}+x)+co{s}^{2}(\frac{2π}{3}-x))}^{'}$

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

$=2\cos x\cdot {(cosx)}^{'}+2cos(\frac{2π}{3}+x)\cdot {(cos(\frac{2π}{3}+x))}^{'}+2cos(\frac{2π}{3}-x)\cdot {(cos(\frac{2π}{3}-x))}^{'}$

$=-2\cos x\cdot \sin x-2\cos (\frac{2π}{3}+x)\sin (\frac{2π}{3}+x)+2\cos (\frac{2π}{3}-x)\sin (\frac{2π}{3}-x)$

$=-\sin 2x-\sin (\frac{4π}{3}+2x)+\sin (\frac{4π}{3}-2x)$

$=-\sin 2x-\sin (π+\frac{π}{3}+2x)+\sin (π+\frac{π}{3}-2x)$

$=-\sin 2x+\sin (\frac{π}{3}+2x)-\sin (\frac{π}{3}-2x)$$=-\sin 2x+2cos\frac{π}{3}\sin 2x$$=-\sin 2x+\sin 2x=0$

Vậy f*'*(x) = 0 với mọi x Î ℝ.