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\(\text{Đặt f (x)= a.cos2x+b.sinx+cosx}\)
\(\text{Hàm f (x) xác định và liên tục trên R}\)
\(\text{f ( π /4 ) = b √2 /2 + √2 /2 }\)
\(\text{f ( 5/π4 ) = − b √ 2/ 2 − √ 2/ 2 }\)
\(\text{⇒ f (π /4) . f ( 5 π/ 4 ) = − 1/2 ( b + 1 )^ 2 ≤ 0 ; ∀ a ; b ; c}\)
\(⇒ f (x)= 0 luôn có ít nhất 1 nghiệm thuộc đoạn [ π /4 ; 5π/4]\)
Hay pt đã có nghiệm.
\(y'=\dfrac{\left(x+\sqrt{x^2+1}\right)'}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{1+\dfrac{x}{\sqrt{x^2+1}}}{2\sqrt{x+\sqrt{x^2+1}}}=\dfrac{x+\sqrt{x^2+1}}{2\sqrt{x^2+1}.\sqrt{x+\sqrt{x^2+1}}}\)
\(=\dfrac{\sqrt{x+\sqrt{x^2+1}}}{2\sqrt{x^2+1}}\)
a: \(y'=4\cdot3x^2-3\cdot2x+2=12x^2-6x+2\)
b: \(y'=\dfrac{\left(x+1\right)'\left(x-1\right)-\left(x+1\right)\left(x-1\right)'}{\left(x-1\right)^2}=\dfrac{x-1-x-1}{\left(x-1\right)^2}=\dfrac{-2}{\left(x-1\right)^2}\)
c: \(y'=-2\cdot\left(\sqrt{x}\cdot x\right)'\)
\(=-2\cdot\left(\dfrac{x+x}{2\sqrt{x}}\right)=-2\cdot\dfrac{2x}{2\sqrt{x}}=-2\sqrt{x}\)
d: \(y'=\left(3sinx+4cosx-tanx\right)\)'
\(=3cosx-4sinx+\dfrac{1}{cos^2x}\)
e: \(y'=\left(4^x+2e^x\right)'\)
\(=4^x\cdot ln4+2\cdot e^x\)
f: \(y'=\left(x\cdot lnx\right)'=lnx+1\)
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 \(y' = \frac{{2{\rm{x}}}}{{{{\sin }^2}\left( {4 - {x^2}} \right)}}\).
\(a,y'=8x^3-9x^2+10x\\ \Rightarrow y''=24x^2-18x+10\\ b,y'=\dfrac{2}{\left(3-x\right)^2}\\ \Rightarrow y''=\dfrac{4}{\left(3-x\right)^3}\)
\(c,y'=2cos2xcosx-sin2xsinx\\ \Rightarrow y''=-5sin\left(2x\right)cos\left(x\right)-4cos\left(2x\right)sin\left(x\right)\\ d,y'=-2e^{-2x+3}\\ \Rightarrow y''=4e^{-2x+3}\)
a.
\(y=\left\{{}\begin{matrix}x-2\left(x\ge2\right)\\2-x\left(x\le2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y'\left(2^+\right)=1\\y'\left(2^-\right)=-1\end{matrix}\right.\)
\(\Rightarrow y'\left(2^+\right)\ne y'\left(2^-\right)\Rightarrow\) không tồn tại đạo hàm tại \(x=2\)
b.
\(y=\left|x-2\right|^2=x^2-4x+4\Rightarrow y'=2x-4\)
\(\Rightarrow y'\left(2\right)=0\)
c.
\(y=\left\{{}\begin{matrix}4-x^2\left(\text{với }-2< x< 2\right)\\x^2-4\left(\text{với }x\ge2;x\le-2\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}y'\left(2^+\right)=2x=4\\y'\left(2^-\right)=-2x=-4\end{matrix}\right.\)
\(\Rightarrow y'\left(2^+\right)\ne y'\left(2^-\right)\Rightarrow\) ko tồn tại đạo hàm tại \(x=2\)
d. Tương tự a và c
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)\(y'=\dfrac{1}{cos^2x}-\dfrac{1}{sin^2x}\)
\(y'=\dfrac{sin^2x-cos^2x}{sin^2x.cos^2x}\)
3.
\(f\left(x+\frac{\pi}{3}\right)=cos\left(x+\frac{\pi}{3}\right)\Rightarrow f'\left(x+\frac{\pi}{3}\right)=-sin\left(x+\frac{\pi}{3}\right)\)
\(f'\left(x-\frac{\pi}{6}\right)=-sin\left(x-\frac{\pi}{6}\right)\)
\(f'\left(0\right)=-sin\left(0\right)=0\)
\(2f'\left(x+\frac{\pi}{3}\right).f'\left(x-\frac{\pi}{6}\right)=2sin\left(x+\frac{\pi}{3}\right)sin\left(x-\frac{\pi}{6}\right)\)
\(=cos\left(\frac{\pi}{2}\right)-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)=0-cos\left(2x+\frac{\pi}{6}\right)=-cos\left(2x+\frac{\pi}{6}\right)\)
\(\Rightarrow2f'\left(x+\frac{\pi}{3}\right)f'\left(x-\frac{\pi}{6}\right)=f'\left(0\right)-f\left(2x+\frac{\pi}{6}\right)\) (đpcm)
4.
\(y=3\left(sin^4x+cos^4x\right)-2\left(sin^6x+cos^6x\right)\)
\(=3\left(sin^2x+cos^2x\right)^2-6sin^2x.cos^2x-2\left(sin^2x+cos^2x\right)^3+6sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)
\(=3-2=1\)
\(\Rightarrow y'=0\) ; \(\forall x\)
5.
\(y=\left(\frac{sinx}{1+cosx}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{1-cos^2x}\right)^3=\left(\frac{sinx\left(1-cosx\right)}{sin^2x}\right)^3=\left(\frac{1-cosx}{sinx}\right)^3\)
\(y'=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{sin^2x-cosx\left(1-cosx\right)}{sin^2x}\right)=3\left(\frac{1-cosx}{sinx}\right)^2\left(\frac{1-cosx}{sin^2x}\right)=\frac{3\left(1-cosx\right)^3}{sin^4x}\)
\(\Rightarrow y'.sinx-3y=\frac{3\left(1-cosx\right)^3}{sin^3x}-3\left(\frac{1-cosx}{sinx}\right)^3=0\) (đpcm)
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) \(y' = {\left( {\sqrt 5 .\sqrt x } \right)^\prime } = \sqrt 5 .\frac{1}{{2\sqrt x }} = \frac{{\sqrt 5 }}{{2\sqrt x }} = \frac{5}{{2\sqrt {5x} }}\).
Ta có:
Chọn D.