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![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(y=x\cdot e^{2x}\)
=>\(y'=\left(x\cdot e^{2x}\right)'\)
\(=x\cdot\left(e^{2x}\right)'+x'\cdot\left(e^{2x}\right)\)
\(=e^{2x}+2\cdot x\cdot e^{2x}\)
\(y''=\left(e^{2x}+2\cdot x\cdot e^{2x}\right)'\)
\(=\left(e^{2x}\right)'+\left(2\cdot x\cdot e^{2x}\right)'\)
\(=4\cdot e^{2x}+4\cdot x\cdot e^{2x}\)
b: \(y=ln\left(2x+3\right)\)
=>\(y'=\dfrac{\left(2x+3\right)'}{\left(2x+3\right)}=\dfrac{2}{2x+3}\)
=>\(y''=\left(\dfrac{2}{2x+3}\right)'=\dfrac{2\left(2x+3\right)'-2'\left(2x+3\right)}{\left(2x+3\right)^2}\)
\(=\dfrac{4}{\left(2x+3\right)^2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
Tính đạo hàm của các hàm số sau:
a) \(y = {x^3} - 3{x^2} + 2x + 1;\)
b) \(y = {x^2} - 4\sqrt x + 3.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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)\(\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}\)
\(\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}\)
\(\dfrac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f'\left(x\right)+g'\left(x\right)\)
\(\dfrac{d}{dx}\left(cf\left(x\right)\right)=cf'\left(x\right)\)
\(y'=\dfrac{d}{dx}\left(x^2\right)-\dfrac{d}{dx}\left(4\sqrt{x}\right)+\dfrac{d}{dx}\left(3\right)\)
\(y'=2x-2\sqrt{x}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(y'=\left(x^2-x\right)'=2x-1\)
\(y''=\left(2x-1\right)'=2\)
b: \(y'=\left(cosx\right)'=-sinx\)
\(y''=\left(-sinx\right)'=-cosx\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,y'=\left(\dfrac{1}{2x+3}\right)'=-\dfrac{2}{\left(2x+3\right)^2}\\ \Rightarrow y''=\dfrac{2\cdot\left[\left(2x+3\right)^2\right]'}{\left(2x+3\right)^4}=\dfrac{8}{\left(2x+3\right)^3}\\ b,y'=\left(log_3x\right)'=\dfrac{1}{xln3}\\ \Rightarrow y''=-\dfrac{1}{x^2ln3}\\ c,y'=\left(2^x\right)'=2^x\cdot ln2\\ \Rightarrow y''=2^x\cdot\left(ln2\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: y=ln(x+1)
=>\(y'=\dfrac{1}{x+1}\)
=>\(y''=\dfrac{1'\left(x+1\right)-1\left(x+1\right)'}{\left(x+1\right)^2}=\dfrac{-1}{\left(x+1\right)^2}\)
b: y=tan 2x
=>\(y'=\dfrac{2}{cos^22x}\)
=>\(y''=\left(\dfrac{2}{cos^22x}\right)'=\dfrac{-2\cdot cos^22x'}{cos^42x}=\dfrac{-2\cdot2\cdot cos2x\left(cos2x\right)'}{cos^42x}\)
\(=\dfrac{4\cdot2\cdot sin2x}{cos^32x}=\dfrac{8\cdot sin2x}{cos^32x}\)
\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)