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1. \(y'=3x^2\sqrt{x}+\dfrac{x^3-5}{2\sqrt{x}}=\dfrac{7x^3-5}{2\sqrt{x}}\)
2. \(y'=3x^5+\dfrac{3}{x^2}+\dfrac{1}{\sqrt{x}}\)
3. \(y'=2-\dfrac{2}{\left(x-2\right)^2}\)
a/ \(y=\left(x^3-3x\right)^{\dfrac{3}{2}}\Rightarrow y'=\dfrac{3}{2}\left(x^3-3x\right)^{\dfrac{1}{2}}\left(x^3-3x\right)'=\dfrac{3}{2}\left(3x^2-3\right)\sqrt{x^3-3x}\)
b/ \(y'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\sqrt{x^3+1}-x^2+2\right)'=5\left(\sqrt{x^3+1}-x^2+2\right)^4\left(\dfrac{3x^2}{\sqrt{x^3+1}}-2x\right)\)c/
\(y'=14\left(x^6+2x-3\right)^6\left(x^6+2x-3\right)'=14\left(x^6+2x-3\right)^6\left(6x^5+2\right)\)
d/ \(y=\left(x^3-1\right)^{-\dfrac{5}{2}}\Rightarrow y'=-\dfrac{5}{2}\left(x^3-1\right)^{-\dfrac{7}{2}}\left(x^3-1\right)'=-\dfrac{15x^2}{2\sqrt{\left(x^3-1\right)^7}}\)
a, \(y=3x^4-7x^3+3x^2+1\)
\(y'=12x^3-21x^2+6x\)
b, \(y=\left(x^2-x\right)^3\)
\(y'=3\left(x^2-x\right)^2\left(2x-1\right)\)
c, \(y=\dfrac{4x-1}{2x+1}\)
\(y'=\dfrac{4+2}{\left(2x+1\right)^2}\)
\(y'=\dfrac{6}{\left(2x+1\right)^2}\)
a: y=3x^4-7x^3+3x^2+1
=>y'=3*4x^3-7*3x^2+3*2x
=12x^3-21x^2+6x
b: \(y'=\left[\left(x^2-x\right)^3\right]'\)
\(=3\left(2x-1\right)\left(x^2-x\right)^2\)
c: \(y'=\dfrac{\left(4x-1\right)'\left(2x+1\right)-\left(4x-1\right)\left(2x+1\right)'}{\left(2x+1\right)^2}\)
\(=\dfrac{4\left(2x+1\right)-2\left(4x-1\right)}{\left(2x+1\right)^2}=\dfrac{6}{\left(2x+1\right)^2}\)
Tính đạo hàm của các hàm số sau:
a) \(y = {x^3} - 3{x^2} + 2x + 1;\)
b) \(y = {x^2} - 4\sqrt x + 3.\)
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) \(y' = 2.3{{\rm{x}}^2} - \frac{1}{2}.2{\rm{x}} + 4.1 - 0 = 6{{\rm{x}}^2} - x + 4\).
b) \(y' = \frac{{{{\left( { - 2{\rm{x}} + 3} \right)}^\prime }.\left( {{\rm{x}} - 4} \right) - \left( { - 2{\rm{x}} + 3} \right).{{\left( {{\rm{x}} - 4} \right)}^\prime }}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
\( = \frac{{ - 2\left( {{\rm{x}} - 4} \right) - \left( { - 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
\( = \frac{{ - 2{\rm{x}} + 8 + 2{\rm{x}} - 3}}{{{{\left( {{\rm{x}} - 4} \right)}^2}}} = \frac{5}{{{{\left( {{\rm{x}} - 4} \right)}^2}}}\)
c) \(y' = \frac{{{{\left( {{x^2} - 2{\rm{x}} + 3} \right)}^\prime }\left( {{\rm{x}} - 1} \right) - \left( {{x^2} - 2{\rm{x}} + 3} \right){{\left( {{\rm{x}} - 1} \right)}^\prime }}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
\( = \frac{{\left( {2{\rm{x}} - 2} \right)\left( {{\rm{x}} - 1} \right) - \left( {{x^2} - 2{\rm{x}} + 3} \right).1}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\) \( = \frac{{2{{\rm{x}}^2} - 2{\rm{x}} - 2{\rm{x}} + 2 - {x^2} + 2{\rm{x}} - 3}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
\( = \frac{{{x^2} - 2{\rm{x}} - 1}}{{{{\left( {{\rm{x}} - 1} \right)}^2}}}\)
d) \(y' = {\left( {\sqrt 5 .\sqrt x } \right)^\prime } = \sqrt 5 .\frac{1}{{2\sqrt x }} = \frac{{\sqrt 5 }}{{2\sqrt x }} = \frac{5}{{2\sqrt {5x} }}\).
\(y'=7\left(-x^2+3x+7\right)^6.\left(-x^2+3x+7\right)'\)
\(=7\left(-2x+3\right)\left(-x^2+3x+7\right)^6\)
\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
Ta có:
Chọn D.