Tính đạo hàm của các hàm số sau:
a) \(y = {x^3} - 3{x^2} + 2x + 1;\)
b) \(y = {x^2} - 4\sqrt x + 3.\)
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NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
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. \(y'=\dfrac{\left(4x-4\right)\left(2x+1\right)-2\left(2x^2-4x+5\right)}{\left(2x+1\right)^2}=\dfrac{4x^2+4x-14}{\left(2x+1\right)^2}\)
NV
Nguyễn Việt Lâm
Giáo viên
30 tháng 4 2021
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. \(y'=4\left(3-2x^2\right)^3.\left(3-2x^2\right)'=-16x\left(3-2x^2\right)^3\)
HQ
Hà Quang Minh
Giáo viên
24 tháng 8 2023
\(a,y'=\left[\left(2x-3\right)^{10}\right]'\\ =10\left(2x-3\right)^9\left(2x-3\right)'\\ =20\left(2x-3\right)^9\\ b,y'=\left(\sqrt{1-x^2}\right)'\\ =\dfrac{\left(1-x^2\right)'}{2\sqrt{1-x^2}}\\ =-\dfrac{2x}{2\sqrt{1-x^2}}\\ =-\dfrac{x}{\sqrt{1-x^2}}\)
B
1
HQ
Hà Quang Minh
Giáo viên
26 tháng 8 2023
\(a,y'=3x^2-4x+2\\ \Rightarrow y''=6x-4\\ b,y'=2xe^x+x^2e^x\\ \Rightarrow y''=4xe^x+x^2e^x+2e^x\)
17 tháng 8 2023
tham khảo:
a)\(y'=\dfrac{\left(2\right)\left(x+2\right)-\left(2x-1\right)\left(1\right)}{\left(x+2\right)^2}\)
\(y'=\dfrac{5}{\left(x+2\right)^2}\)
b)\(y'=\dfrac{\left(2\right)\left(x^2+1\right)-\left(2x\right)\left(2x\right)}{\left(x^2+1\right)^2}\)
\(y'=\dfrac{2\left(1-x^2\right)}{\left(x^2+1\right)^2}\)
17 tháng 8 2023
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: \(y=ln\left(2x+3\right)\)
=>\(y'=\dfrac{\left(2x+3\right)'}{\left(2x+3\right)}=\dfrac{2}{2x+3}\)
=>\(y''=\left(\dfrac{2}{2x+3}\right)'=\dfrac{2\left(2x+3\right)'-2'\left(2x+3\right)}{\left(2x+3\right)^2}\)
\(=\dfrac{4}{\left(2x+3\right)^2}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) Đặt \(u = 3{\rm{x}}\) thì \(y = \sin u\). Ta có: \(u{'_x} = {\left( {3{\rm{x}}} \right)^\prime } = 3\) và \(y{'_u} = {\left( {\sin u} \right)^\prime } = \cos u\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = \cos u.3 = 3\cos 3{\rm{x}}\).
Vậy \(y' = 3\cos 3{\rm{x}}\).
b) Đặt \(u = \cos 2{\rm{x}}\) thì \(y = {u^3}\). Ta có: \(u{'_x} = {\left( {\cos 2{\rm{x}}} \right)^\prime } = - 2\sin 2{\rm{x}}\) và \(y{'_u} = {\left( {{u^3}} \right)^\prime } = 3{u^2}\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = 3{u^2}.\left( { - 2\sin 2{\rm{x}}} \right) = 3{\left( {\cos 2{\rm{x}}} \right)^2}.\left( { - 2\sin 2{\rm{x}}} \right) = - 6\sin 2{\rm{x}}{\cos ^2}2{\rm{x}}\).
Vậy \(y' = - 6\sin 2{\rm{x}}{\cos ^2}2{\rm{x}}\).
c) Đặt \(u = \tan {\rm{x}}\) thì \(y = {u^2}\). Ta có: \(u{'_x} = {\left( {\tan {\rm{x}}} \right)^\prime } = \frac{1}{{{{\cos }^2}x}}\) và \(y{'_u} = {\left( {{u^2}} \right)^\prime } = 2u\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = 2u.\frac{1}{{{{\cos }^2}x}} = 2\tan x\left( {{{\tan }^2}x + 1} \right)\).
Vậy \(y' = 2\tan x\left( {{{\tan }^2}x + 1} \right)\).
d) Đặt \(u = 4 - {x^2}\) thì \(y = \cot u\). Ta có: \(u{'_x} = {\left( {4 - {x^2}} \right)^\prime } = - 2{\rm{x}}\) và \(y{'_u} = {\left( {\cot u} \right)^\prime } = - \frac{1}{{{{\sin }^2}u}}\).
Suy ra \(y{'_x} = y{'_u}.u{'_x} = - \frac{1}{{{{\sin }^2}u}}.\left( { - 2{\rm{x}}} \right) = \frac{{2{\rm{x}}}}{{{{\sin }^2}\left( {4 - {x^2}} \right)}}\).
Vậy \(y' = \frac{{2{\rm{x}}}}{{{{\sin }^2}\left( {4 - {x^2}} \right)}}\).
17 tháng 8 2023
tham khảo:
a)\(y'=xsin2x+sin^2x\)
\(y'=sin^2x+xsin2x\)
b)\(y'=-2sin2x+2cosx\\ y'=2\left(cosx-sin2x\right)\)
c)\(y=sin3x-3sinx\)
\(y'=3cos3x-3cosx\)
d)\(y'=\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\)
\(y'=\dfrac{sin^2x-cos^2x}{sin^2x.cos^2x}\)
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)\(\dfrac{d}{dx}\left(x^n\right)=nx^{n-1}\)
\(\dfrac{d}{dx}\left(\sqrt{x}\right)=\dfrac{1}{2\sqrt{x}}\)
\(\dfrac{d}{dx}\left(f\left(x\right)+g\left(x\right)\right)=f'\left(x\right)+g'\left(x\right)\)
\(\dfrac{d}{dx}\left(cf\left(x\right)\right)=cf'\left(x\right)\)
\(y'=\dfrac{d}{dx}\left(x^2\right)-\dfrac{d}{dx}\left(4\sqrt{x}\right)+\dfrac{d}{dx}\left(3\right)\)
\(y'=2x-2\sqrt{x}\)