https://drive.google.com/file/d/14Q-YI3szy-rePnIHWGD35RKCWiCXCT6k/view?usp=sharing

Võ Ngọc Tú Uyên  

a. \(y'=\dfrac{-1}{\left(x-1\right)}\)

b. \(y'=\dfrac{5}{\left(1-3x\right)^2}\)

c. \(y=\dfrac{\left(x+1\right)^2+1}{x+1}=x+1+\dfrac{1}{x+1}\Rightarrow y'=1-\dfrac{1}{\left(x+1\right)^2}=\dfrac{x^2+2x}{\left(x+1\right)^2}\)

d. \(y'=\dfrac{4x\left(x^2-2x-3\right)-2x^2\left(2x-2\right)}{\left(x^2-2x-3\right)^2}=\dfrac{-4x^2-12x}{\left(x^2-2x-3\right)^2}\)

e. \(y'=1+\dfrac{2}{\left(x-1\right)^2}=\dfrac{x^2-2x+3}{\left(x-1\right)^2}\)

g. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)

2.

a. \(y'=4\left(x^2+x+1\right)^3.\left(x^2+x+1\right)'=4\left(x^2+x+1\right)^3\left(2x+1\right)\)

b. \(y'=5\left(1-2x^2\right)^4.\left(1-2x^2\right)'=-20x\left(1-2x^2\right)^4\)

c. \(y'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{2x+1}{x-1}\right)'=3\left(\dfrac{2x+1}{x-1}\right)^2.\left(\dfrac{-3}{\left(x-1\right)^2}\right)=\dfrac{-9\left(2x+1\right)^2}{\left(x-1\right)^4}\)

d. \(y'=\dfrac{2\left(x+1\right)\left(x-1\right)^3-3\left(x-1\right)^2\left(x+1\right)^2}{\left(x-1\right)^6}=\dfrac{-x^2-6x-5}{\left(x-1\right)^4}\)

e. \(y'=-\dfrac{\left[\left(x^2-2x+5\right)^2\right]'}{\left(x^2-2x+5\right)^4}=-\dfrac{2\left(x^2-2x+5\right)\left(2x-2\right)}{\left(x^2-2x+5\right)^4}=-\dfrac{4\left(x-1\right)}{\left(x^2-2x+5\right)^3}\)

f. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)

a) Với bất kì \({x_0} \in \mathbb{R}\), ta có:

\(f'\left( {{x_0}} \right) = \mathop {\lim }\limits_{x \to {x_0}} \frac{{f\left( x \right) - f\left( {{x_0}} \right)}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} \frac{{x - {x_0}}}{{x - {x_0}}} = \mathop {\lim }\limits_{x \to {x_0}} 1 = 1\)

Vậy \(f'\left( x \right) = {\left( x \right)^\prime } = 1\) trên \(\mathbb{R}\).

b) Ta có:

\(\begin{array}{l}{\left( {{x^2}} \right)^\prime } = 2{\rm{x}}\\{\left( {{x^3}} \right)^\prime } = 3{{\rm{x}}^2}\\...\\{\left( {{x^n}} \right)^\prime } = n{{\rm{x}}^{n - 1}}\end{array}\)

\(a,y'=\left(f\left(g\left(x\right)\right)\right)'\)

\(=f'\left(g\left(x\right)\right).g'\left(x\right)\)

\(=e^{g\left(x\right)}.\left(2x-1\right)\)

\(=e^{x^2-x}.\left(2x-1\right)\)

\(b,y'=\dfrac{d}{dx}\left(3^{sinx}\right)\)

\(=\dfrac{d}{dx}\left(e^{ln3.sinx}\right)\)

\(=\dfrac{d}{dx}\left(ln3.sinx\right).e^{ln3.sinx}\)

\(=ln3.cosx.3^{sinx}\)

Chọn C.

Đầu tiên sử dụng quy tắc nhân.

y' = [(x² – x + 1)³]’(x² + x + 1)² + [(x² + x + 1)²]’(x² – x + 1)³.

Sau đó sử dụng công thức 

y' = 3(x² – x + 1)²(x² – x + 1)’(x² + x  + 1) + 2(x² + x + 1)(x² + x + 1)’(x² – x + 1)³

y’ = 3(x² – x + 1)²(2x – 1)(x² + x + 1)² + 2(x² + x + 1)(2x + 1)(x² – x + 1)³

y’ = (x² – x + 1)²(x² + x + 1)[3(2x – 1)(x² + x + 1) + 2(2x + 1)(x² – x + 1)].

tham khảo:

a)\(y'=\dfrac{d}{dx}\left(x^3\right)-\dfrac{d}{dx}\left(3x^2\right)+\dfrac{d}{dx}\left(2x\right)+\dfrac{d}{dx}\left(1\right)\)

\(y'=3x^2-6x+2\)

b)\(\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}\)

\(\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}\)

\(\dfrac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f'\left(x\right)+g'\left(x\right)\)

\(\dfrac{d}{dx}\left(cf\left(x\right)\right)=cf'\left(x\right)\)

\(y'=\dfrac{d}{dx}\left(x^2\right)-\dfrac{d}{dx}\left(4\sqrt{x}\right)+\dfrac{d}{dx}\left(3\right)\)

\(y'=2x-2\sqrt{x}\)

\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)

Tổng quát: Đạo hàm cấp n của y: