Tính đạo hàm của các hàm số sau :
\(a,y=log^2\left(2x+1\right)\)
\(b,y=\frac{x+1}{4^x}\)
\(c,y=13x\)
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17 tháng 8 2023
tham khảo:
a)\(y'\left(x\right)=5\left(\dfrac{2x-1}{x+2}\right)^4.\dfrac{\left(x+2\right)\left(2\right)-\left(2x-1\right).1}{\left(x+2\right)^2}\)
\(=\dfrac{10\left(2x-1\right)\left(x+2\right)^3}{\left(x+2\right)^4}=\dfrac{20x-50}{\left(x+2\right)^4}\)
b)\(y'\left(x\right)=\dfrac{2\left(x^2+1\right)-2x\left(2x\right)}{\left(x^2+1\right)^2}\)\(=\dfrac{2\left(1-x^2\right)}{\left(x^2+1\right)^2}\)
c)\(y'\left(x\right)=e^x.2sinxcosx+e^xsin^2x.2cosx\)
\(=2e^xsinx\left(cosx+sinxcosx\right)\)
\(=2e^xsinxcos^2x\)
d)\(y'\left(x\right)=\dfrac{1}{x\sqrt{x}}.\left(+\dfrac{1}{2\sqrt{x}}\right)\)
\(=\dfrac{1}{\sqrt{x}\left(2\sqrt{x}+\sqrt{x}+2\right)}\)
\(=\dfrac{1}{\sqrt{x}\left(3\sqrt{x}+2\right)}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 11 2021
a.
\(y'=\dfrac{\left(1+\sqrt{3x-1}\right)'}{1+\sqrt{3x-1}}=\dfrac{3}{2\left(1+\sqrt{3x-1}\right)\sqrt{3x-1}}\)
b.
\(y'=\dfrac{\left(2sin^2x-1\right)'}{\left(2sin^2x-1\right).ln10}=\dfrac{2sin2x}{\left(2sin^2x-1\right)ln10}\)
c.
\(y'=\left(3x^2+3\right)3^{x^3+3x+1}.e^x.ln3+3^{x^3+3x+1}.e^x\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
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 \(y' = 3{{\rm{e}}^{3{\rm{x}} + 1}}\).
20 tháng 8 2023
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: \(y'=\left(x^3\right)'\cdot log_2x+x^3\cdot\left(log_2x\right)'\)
\(=3x^2\cdot log_2x+x^3\cdot\dfrac{1}{x\cdot ln2}\)
17 tháng 8 2023
tham khảo:
a)y′=2\(^{3x-x^2}\).ln2.(3−2x)
b) y′\(\dfrac{4}{ln3}\).\(\dfrac{1}{4x+1}\).4=\(\dfrac{4}{\left(4x+1\right)ln3}\)
20 tháng 8 2023
a, \(y=\left(2x^3+3\right)^2\)
\(y'=2\left(2x^3+3\right)6x^2\)
\(=12x^2\left(2x^3+3\right)\)
b,\(y=cos3x\)
\(y'=-3sin3x\)
c, \(y=log_2\left(x^2+2\right)\)
\(y'=\dfrac{2x}{\left(x^2+2\right)ln2}\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 8 2023
\(a,D=R\\ b,2x-3>0\\ \Rightarrow x>\dfrac{3}{2}\\ \Rightarrow D=(\dfrac{3}{2};+\infty)\\ c,-x^2+4>0\\ \Rightarrow x^2< 4\\ \Leftrightarrow-2< x< 2\\ \Rightarrow D=\left(-2;2\right)\)
HQ
Hà Quang Minh
Giáo viên
22 tháng 9 2023
a) \(g'\left( x \right) = y' = {\left( {2x + \frac{\pi }{4}} \right)^,}.\cos \left( {2x + \frac{\pi }{4}} \right) = 2\cos \left( {2x + \frac{\pi }{4}} \right)\)
b) \(g'\left( x \right) = - 2{\left( {2x + \frac{\pi }{4}} \right)^,}.\sin \left( {2x + \frac{\pi }{4}} \right) = - 4\sin \left( {2x + \frac{\pi }{4}} \right)\)
25 tháng 11 2023
a: \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
=>\(y'=\dfrac{1}{3}\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}\cdot\left(2x^2-x+1\right)'\)
\(=\dfrac{1}{3}\cdot\left(4x-1\right)\left(2x^2-x+1\right)^{-\dfrac{2}{3}}\)
b: \(y=\left(3x+1\right)^{\Omega}\)
=>\(y'=\Omega\cdot\left(3x+1\right)'\cdot\left(3x+1\right)^{\Omega-1}\)
=>\(y'=3\Omega\left(3x+1\right)^{\Omega-1}\)
c: \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
=>\(y'=\dfrac{\left(\dfrac{1}{x-1}\right)'}{3\cdot\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(=\dfrac{\dfrac{1'\left(x-1\right)-\left(x-1\right)'\cdot1}{\left(x-1\right)^2}}{\dfrac{3}{\sqrt[3]{\left(x-1\right)^2}}}\)
\(=\dfrac{-x}{\left(x-1\right)^2}\cdot\dfrac{\sqrt[3]{\left(x-1\right)^2}}{3}\)
\(=\dfrac{-x}{\sqrt[3]{\left(x-1\right)^4}\cdot3}\)
d: \(y=log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Leftrightarrow y'=\dfrac{\left(\dfrac{x+1}{x-1}\right)'}{\dfrac{x+1}{x-1}\cdot ln3}\)
\(\Leftrightarrow y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}:\dfrac{ln3\left(x+1\right)}{x-1}\)
\(\Leftrightarrow y'=\dfrac{x-1-x-1}{\left(x-1\right)^2}\cdot\dfrac{x-1}{ln3\cdot\left(x+1\right)}\)
\(\Leftrightarrow y'=\dfrac{-2}{\left(x-1\right)\cdot\left(x+1\right)\cdot ln3}\)
e: \(y=3^{x^2}\)
=>\(y'=\left(x^2\right)'\cdot ln3\cdot3^{x^2}=2x\cdot ln3\cdot3^{x^2}\)
f: \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
=>\(y'=\left(x^2-1\right)'\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}=2x\cdot ln\left(\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)^{x^2-1}\)
h: \(y=\left(x+1\right)\cdot e^{cosx}\)
=>\(y'=\left(x+1\right)'\cdot e^{cosx}+\left(x+1\right)\cdot\left(e^{cosx}\right)'\)
=>\(y'=e^{cosx}+\left(x+1\right)\cdot\left(cosx\right)'\cdot e^u\)
\(=e^{cosx}+\left(x+1\right)\cdot\left(-sinx\right)\cdot e^u\)
25 tháng 11 2023
a) \(y=\left(2x^2-x+1\right)^{\dfrac{1}{3}}\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{\dfrac{1}{3}-1}.\left(4x-1\right)\)
\(\Rightarrow y'=\dfrac{1}{3}.\left(2x^2-x+1\right)^{-\dfrac{2}{3}}.\left(4x-1\right)\)
b) \(y=\left(3x+1\right)^{\pi}\)
\(\Rightarrow y'=\pi.\left(3x+1\right)^{\pi-1}.3=3\pi.\left(3x+1\right)^{\pi-1}\)
c) \(y=\sqrt[3]{\dfrac{1}{x-1}}\)
\(\Rightarrow y'=\dfrac{\left(x-1\right)^{-1-1}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^{3-1}}}=\dfrac{\left(x-1\right)^{-2}}{3\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}=\dfrac{1}{3.\sqrt[]{x-1}.\sqrt[3]{\left(\dfrac{1}{x-1}\right)^2}}\)
\(\Rightarrow y'=\dfrac{1}{3\left(x-1\right)^{\dfrac{1}{2}}.\left(x-1\right)^{\dfrac{2}{3}}}=\dfrac{1}{3\left(x-1\right)^{\dfrac{7}{6}}}=\dfrac{1}{3\sqrt[6]{\left(x-1\right)^7}}\)
d) \(y=\log_3\left(\dfrac{x+1}{x-1}\right)\)
\(\Rightarrow y'=\dfrac{\dfrac{1-\left(-1\right)}{\left(x-1\right)^2}}{\dfrac{x+1}{x-1}.\ln3}=\dfrac{2}{\left(x+1\right)\left(x-1\right).\ln3}\)
e) \(y=3^{x^2}\)
\(\Rightarrow y'=3^{x^2}.ln3.2x=2x.3^{x^2}.ln3\)
f) \(y=\left(\dfrac{1}{2}\right)^{x^2-1}\)
\(\Rightarrow y'=\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}.2x=2x.\left(\dfrac{1}{2}\right)^{x^2-1}.ln\dfrac{1}{2}\)
Các bài còn lại bạn tự làm nhé!
HQ
Hà Quang Minh
Giáo viên
22 tháng 8 2023
\(a,y'=\left(\dfrac{1}{2x+3}\right)'=-\dfrac{2}{\left(2x+3\right)^2}\\ \Rightarrow y''=\dfrac{2\cdot\left[\left(2x+3\right)^2\right]'}{\left(2x+3\right)^4}=\dfrac{8}{\left(2x+3\right)^3}\\ b,y'=\left(log_3x\right)'=\dfrac{1}{xln3}\\ \Rightarrow y''=-\dfrac{1}{x^2ln3}\\ c,y'=\left(2^x\right)'=2^x\cdot ln2\\ \Rightarrow y''=2^x\cdot\left(ln2\right)^2\)