a) Cách 1: y' = (9 -2x)'(2x3- 9x2 +1) +(9 -2x)(2x3- 9x2 +1)' = -2(2x3- 9x2 +1) +(9 -2x)(6x2 -18x) = -16x3 +108x2 -162x -2.
Cách 2: y = -4x4 +36x3 -81x2 -2x +9, do đó
y' = -16x3 +108x2 -162x -2.
b) y' = .(7x -3) +
(7x -3)'=
(7x -3) +7
.
c) y' = (x -2)'√(x2 +1) + (x -2)(√x2 +1)' = √(x2 +1) + (x -2) = √(x2 +1) + (x -2)
= √(x2 +1) +
=
.
d) y' = 2tanx.(tanx)' - (x2)' =
.
e) y' = sin
=
sin
.
a: \(y'=\left[tan\left(e^x+1\right)\right]'=\dfrac{\left(e^x+1\right)'}{cos^2\left(e^x+1\right)}=\dfrac{e^x}{cos^2\left(e^x+1\right)}\)
b: \(y'=\left(\sqrt{sin3x}\right)'\)
\(=\dfrac{\left(sin3x\right)'}{2\sqrt{sin3x}}=\dfrac{3\cdot cos3x}{2\sqrt{sin3x}}\)
c: \(y=cot\left(1-2^x\right)\)
=>\(y'=\left[cot\left(1-2^x\right)\right]'\)
\(=\dfrac{-2}{sin^2\left(1-2^x\right)}\cdot\left(-2^x\cdot ln2\right)\)
\(=\dfrac{2^{x+1}\cdot ln2}{sin^2\left(1-2^x\right)}\)
\(a,y=\left(u\left(x\right)\right)^2=\left(x^2+1\right)^2=x^4+2x^2+1\\ b,y'\left(x\right)=4x^3+4x,u'\left(x\right)=2x,y'\left(u\right)=2u\\ \Rightarrow y'\left(u\right)\cdot u'\left(x\right)=2u\cdot2x=4x\left(x^2+1\right)=4x^3+4x\)
Vậy \(y'\left(x\right)=y'\left(u\right)\cdot u'\left(x\right)\)
\(y=\dfrac{x+3}{x+2}\)
=>\(y'=\dfrac{\left(x+3\right)'\left(x+2\right)-\left(x+3\right)\left(x+2\right)'}{\left(x+2\right)^2}=\dfrac{x+2-x-3}{\left(x+2\right)^2}=\dfrac{-1}{\left(x+2\right)^2}\)
=>C
HD: áp dụng BĐT Cô-si cho 3 số hạng trên, khi đó trong căn sẽ triệt tiêu các tổng suy ra đpcm
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
\(y'=\tan x+\frac{x}{\cos^2x}\)
\(y''=\frac{1}{\cos^2x}+\frac{\cos^2-x.2\cos x.\left(-\sin x\right)}{\cos^4x}=\frac{2\cos^2x+2x.\sin x.\cos x}{\cos^4x}\)
\(VT=\frac{2x^2\left(\cos^2x+x\sin x.\cos x\right)}{\cos^4x}\)
\(VP=2\left(x^2+x^2\tan^2x\right)\left(1+x\tan x\right)\)
\(=\frac{2x^2\left(1+x\tan x\right)}{\cos^2x}=\frac{2x^2\left(\cos^2x+x\sin x.\cos x\right)}{\cos^4x}=VT\)
Ta có y ′ = tan x + x ( 1 + tan 2 x ) y ′ = tan x + x ( 1 + tan 2 x ) ⇒ y ′′ = 1 + tan 2 x + 1 + tan 2 x + 2 x tan x ( 1 + tan 2 x ) = 2 ( 1 + tan 2 x ) ( 1 + x tan x ) ⇒ y ″ = 1 + tan 2 x + 1 + tan 2 x + 2 x tan x ( 1 + tan 2 x ) = 2 ( 1 + tan 2 x ) ( 1 + x tan x ) ⇒ x 2 y ′′ = 2 ( x 2 + x 2 tan 2 x ) ( 1 + x tan x ) = 2 ( x 2 + y 2 ) ( 1 + y ) ⇒ x 2 y ″ = 2 ( x 2 + x 2 tan 2 x ) ( 1 + x tan x ) = 2 ( x 2 + y 2 ) ( 1 + y ) ( đpcm )
y'=(xtanx)' = tanx + x(tanx)' = tanx + x\(\dfrac{1}{\cos^2x}\)
=\(\tan x+x\left(1+\tan^2x\right)\)
y''=\(1+\tan^2x+1+\tan^2x+2x\tan x\left(1+\tan^2x\right)=2\left(1+\tan^2x\right)\left(1+x\tan x\right)\)
\(x^2y''=2\left(x^2+x^2\tan^2x\right)\left(1+x\tan x\right)=2\left(x^2+y^2\right)\left(1+y\right)\)
<br class="Apple-interchange-newline"><div></div>y'=tanx+xcos2x
y''=1cos2x +cos2−x.2cosx.(−sinx)cos4x =2cos2x+2x.sinx.cosxcos4x
VT=2x2(cos2x+xsinx.cosx)cos4x
VP=2(x2+x2tan2x)(1+xtanx)
=2x2(1+xtanx)cos2x =2x2(cos2x+xsinx.cosx)cos4x
Ta có y′=tanx+x(1+tan2x)y′=tanx+x(1+tan2x)
⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)⇒y″=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
⇒x2y′′=2(x2+x2tan
Ta có y′=tanx+x(1+tan2x)y′=tanx+x(1+tan2x)
⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)⇒y″=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
⇒x2y′′=2(x2+x2tan
Ta có y′=tanx+x(1+tan2x)y′=tanx+x(1+tan2x)
⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)⇒y″=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
⇒x2y′′=2(x2+x2tan
Ta có y'=tanx+x(1+tan²x)
=>Y mũ n = 1+tan²x+1+tan²x +2x tanx (1+tan²x)=2(1+tan²x) (1+xtanx)
=> X²(y mũ n) =2(x²+x²tan²x )(1+xtanx)=2(x²+y²)(1+y)(đpcm)
ta có y' = tan x +x.( 1+tan 2x)
⇒y'' =1 + tan2x +1 +tan2x +2.x.tan x.(1+tan2x ) 9=) 2+ 2.tan2x +2.x.tan x.(1+tan2x )
(=) 2. (1 +tan2x )+2.x.tan x.(1 +tan2x ) (=) 2.(1 +tan2x ).(1+xtanx)
xét vế trái ta có : x2y'' = x2.2.(1 +tan2x ).(1+xtanx) (=) 2.(1+xtanx).(x2 +x2.tan2x) (1)
theo đề bài ta có y = xtanx
⇒ y2 = x2.tan2x
⇒ (1) (=) : 2.(x
2+y2
).(1+y)
Ta có {y}'=\tan x+x\left( 1+{{\tan }^{2}}x \right)y′=tanx+x(1+tan2x)
\Rightarrow y''=1+{{\tan }^{2}}x+1+{{\tan }^{2}}x+2x\tan x\left( 1+{{\tan }^{2}}x \right)=2\left( 1+{{\tan }^{2}}x \right)\left( 1+x\tan x \right)⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
\Rightarrow {{x}^{2}}y''=2\left( {{x}^{2}}+{{x}^{2}}{{\tan }^{2}}x \right)\left( 1+x\tan x \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)\left( 1+y \right)⇒x2y′′=2(x2+x2tan2x)(1+xtanx)=2(x2+y2)(1+y) ( đpcm ).
Ta có {y}'=\tan x+x\left( 1+{{\tan }^{2}}x \right)y′=tanx+x(1+tan2x)
\Rightarrow y''=1+{{\tan }^{2}}x+1+{{\tan }^{2}}x+2x\tan x\left( 1+{{\tan }^{2}}x \right)=2\left( 1+{{\tan }^{2}}x \right)\left( 1+x\tan x \right)⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
\Rightarrow {{x}^{2}}y''=2\left( {{x}^{2}}+{{x}^{2}}{{\tan }^{2}}x \right)\left( 1+x\tan x \right)=2\left( {{x}^{2}}+{{y}^{2}} \right)\left( 1+y \right)⇒x2y′′=2(x2+x2tan2x)(1+xtanx)=2(x2+y2)(1+y) ( đpcm ).
Ta có y'=tan x+x*[1+(tanx)^2]
⇒ y''=1+(tanx)^2+1+(tanx)^2+2*x*tanx(1+(tanx)^2 =2*(1+(tanx)^2)*( 1+x*tanx)
(x^2)*y''=2*[ x^2+(x^2)*(tanx)^2 ]*(1+x*tan x)=2*(x^2+y^2)*(1+y)( đpcm ).
2Ta có {y}'=\tan x+x\left( 1+{{\tan }^{2}}x \right)y
'=tan x+x (1+tan2
x)
=>y"=1+tan2x+1+tan2x+2x tan x
⇒x2y'' =2(x2+x2tan2x) (1+xtan x) =2(x2+y2) (1+y) ( đpcm)
Ta có y′=tanx+x(1+tan2x)y′=tanx+x(1+tan2x)
⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)⇒y″=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
⇒x2y′′=2(x2+x2tan
Ta có y′=tanx+x(1+tan2x)y′=tanx+x(1+tan2x)
⇒y′′=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)⇒y″=1+tan2x+1+tan2x+2xtanx(1+tan2x)=2(1+tan2x)(1+xtanx)
⇒x2y′′=2(x2+x2tan
y’=tan x+x(1+tan2x)
=>y’’=1+tan2x+1+tan2x+ 2xtanx(1+tan2x)= 2(1+tan2x)(1+xtanx)
=> x2y’’ = 2(x2 + x2tan2x)(1+xtanx) =2(x2+y2)(1+y) (ĐPCM)