Tính đạo hàm của các hàm số sau: y = e x - e - x x
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a: \(y'=\left(x\cdot log_2x\right)'=log_2x+x\cdot\dfrac{1}{x\cdot ln2}=log_2x+\dfrac{1}{ln2}\)
b: \(y'=\left(x^3e^x\right)'=\left(x^3\right)'\cdot e^x+x^3\cdot\left(e^x\right)'\)
\(=3x^2\cdot e^x+x^3\cdot e^x\)
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\(a,y'=\left(f\left(g\left(x\right)\right)\right)'\)
\(=f'\left(g\left(x\right)\right).g'\left(x\right)\)
\(=e^{g\left(x\right)}.\left(2x-1\right)\)
\(=e^{x^2-x}.\left(2x-1\right)\)
\(b,y'=\dfrac{d}{dx}\left(3^{sinx}\right)\)
\(=\dfrac{d}{dx}\left(e^{ln3.sinx}\right)\)
\(=\dfrac{d}{dx}\left(ln3.sinx\right).e^{ln3.sinx}\)
\(=ln3.cosx.3^{sinx}\)
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\(a,y'=8x^3-10x\\ \Rightarrow y''=24x^2-10\\ b,y'=e^x+xe^x\\ \Rightarrow y''=e^x+e^x+xe^x=2e^x+xe^x\)
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a: \(y'=\left(x^2+3x-1\right)'\cdot e^x+\left(x^2+3x-1\right)\cdot\left(e^x\right)'\)
\(=e^x\left(2x+3\right)+\left(x^2+3x-1\right)\cdot e^x\)
\(=e^x\left(x^2+5x+2\right)\)
b: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)
\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)
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\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)
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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. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(y' = {\left( {{x^2} - x} \right)^\prime }{.2^x} + \left( {{x^2} - x} \right).{\left( {{2^x}} \right)^\prime } = \left( {2{\rm{x}} - 1} \right){.2^x} + \left( {{x^2} - x} \right){.2^x}.\ln 2\).
b) \(y' = {\left( {{x^2}} \right)^\prime }.{\log _3}x + {x^2}.{\left( {{{\log }_3}x} \right)^\prime } = 2{\rm{x}}.{\log _3}x + {x^2}.\frac{1}{{x\ln 3}} = 2{\rm{x}}.{\log _3}x + \frac{x}{{\ln 3}}\).
c) Đặt \(u = 3{\rm{x}} + 1\) thì \(y = {e^u}\). Ta có: \(u{'_x} = {\left( {3{\rm{x}} + 1} \right)^\prime } = 3\) và \(y{'_u} = {\left( {{e^u}} \right)^\prime } = {e^u}\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = {e^u}.3 = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).
Vậy \(y' = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).
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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: \(y=ln\left(2x+3\right)\)
=>\(y'=\dfrac{\left(2x+3\right)'}{\left(2x+3\right)}=\dfrac{2}{2x+3}\)
=>\(y''=\left(\dfrac{2}{2x+3}\right)'=\dfrac{2\left(2x+3\right)'-2'\left(2x+3\right)}{\left(2x+3\right)^2}\)
\(=\dfrac{4}{\left(2x+3\right)^2}\)