alimov-11-2003-gdz- (Алгебра - 10-11 класс - Алимов), страница 2

(Опечатка в ответе задачника).№ 819.()() ( x )′ =′′22⎛ x2 − 4 ⎞⎟ = x −4 x − x −4 ⋅1) ⎜2⎜ x ⎟⎠⎝x()22x x − x 2 − 4 ⋅( )1222x = 4 x − x + 4 = 3x + 4 ;x2x x2x x′′⎛⎛1 ⎞⎛1 ⎞⎞ ⎛1 ⎞11x +1⎟⎟ =.2) ⎜ ⎜⎜ 4 x + 4 ⎟⎟⎜⎜ 4 x − 4 ⎟⎟ ⎟ = ⎜⎜ x −+=⎜⎟x ⎠⎝x ⎠⎠ ⎝x⎠2 x 2x x 2x x⎝⎝(Опечатка в ответе задачника).=№ 820.((()()))′1) (2x − 3)5 3x 2 + 2 x + 1 =′′= (2 x − 3)5 3x 2 + 2 x + 1 + (2x − 3)5 3x 2 + 2 x + 1 =(())= 5 ⋅ 2(2 x − 3) 3x + 2 x + 1 + (2x − 3) (6 x + 2 ) =4(25)()= (2x − 3) 30x + 20x + 10 + 12x 2 + 4x − 18x − 6 = (2 x − 3) 42 x 2 + 6 x + 4 ;′′′2) (x − 1)4 (x + 1)7 = (x − 1)4 (x + 1)7 + (x − 1)4 (x + 1)7 =4(2) ()4()= 4(x − 1) (x + 1) + 7(x − 1) (x + 1) = (x − 1) (x + 1) ( 4x + 4 + 7 x − 7) =374636= (x − 1)3 (x + 1)6 (11x − 3) ;′′′3) 4 3x + 2 (3x − 1)4 = 4 3x + 2 (3x − 1)4 + 4 3x + 2 (3x − 1)4 =(==4)===) (3 ⋅ (3x − 1)444 (3x + 2)3()+ 4 3x + 2 ⋅ 4 ⋅ 3 ⋅ (3x − 1)3 =)3(3x − 1) + 48(3x + 2)(3x − 1)3444 (3x + 2)3(3x − 1)3 (9x − 3 + 144x + 96) = 3(3x − 1)3 (51x + 31) ;44 (3x + 2 )344 (3x + 2 )3( 2 x + 1 ⋅ (2 x − 3) )′ = ( 2x + 1 )′ (2x − 3) +332(2 x − 3)33 (2 x + 1)322(2 x − 3)233 (2 x + 1)2333(=)′2 x + 1 (2 x − 3)3 =+ 3 2 x + 1 ⋅ 2 ⋅ 3(2 x − 3)2 =(2 x − 3 + 18 x + 9) =4(2 x − 3)2 (10 x + 3)33 (2 x + 1)22(2 x − 3)233 (2 x + 1)2( 20 x + 6) =.15№ 821.()()′′′22⎛ 2 x 2 − 3x + 1 ⎞⎟ = 2 x − 3x + 1 (x + 1) − 2 x − 3x + 1 (x + 1) =1) ⎜2⎜⎟x +1(x + 1)⎝⎠==(4x − 3)(x + 1) − (2x 2 − 3x + 1) =(x + 1)24 x 2 + 4 x − 3x − 3 − 2 x 2 + 3x − 1(x + 1)2(2x 2 + 4x − 4=(x + 1)2);()′′′22⎛ 3x 2 + 2 x − 1 ⎞⎟ = 3x + 2 x − 1 (2 x + 1) − 3x + 2 x − 1 (2x + 1) =2) ⎜2⎜ 2x + 1 ⎟(2x + 1)⎝⎠==3)(6x + 2)(2x + 1) − 2(3x 2 + 2x − 1) =(2x + 1)212 x 2 + 6 x + 4 x + 2 − 6x 2 − 4 x + 2(2x + 1)22(2 − x )2 + ( x )2−xx+=2−x(2 − x ) xx6x 2 + 6x + 4==(2x + 1)2;4 − 4x + x 2 + x=x 2 − 3x + 42 x −x x2 x −x x′′′22⎛ x 2 − 3x + 4 ⎞⎜⎟ = = x − 3x + 4 2 x − x x − x − 3x + 4 2 x − x x =2⎜2 x −x x ⎟⎝⎠2 x −x x(===(2x − 3)(2)()()x − x x − x 2 − 3x + 4 ⎛⎜⎝(2x −x x))((1x−232)()x ⎞⎟⎠=4x x − 2x 2 x − 6 x + 3x x − x x + 32 x 2 x + 3 x − 92 x x −(2− 12 x 2 x + 32 x x + 3 x −(2x −x x)24xx −x x=)− x 3 + 3x 2 + 6x − 82 x x ⋅ (2 − x )2+6 x.f ′(x)=(2x3–3x2–12x+1)′=6x2–6x–12; f ′(0)=6x2–6x–12=0;1+ 31− 3D=1+8=9;х1==2, х2== –1.22№ 823.4x2№ 822.′′′⎛ 2 x − 1 ⎞ (2 x − 1) (x + 1) − (2 x − 1)(x + 1)f ′(x)= ⎜=⎟ =⎝ x +1 ⎠(x + 1)216)x2–x–2=0;==2( x + 1) − (2 x − 1)(x + 1)2=2x + 2 − 2x + 1(x + 1)2(х+1)2=1;f ′(x)=3 ⇒=x2+x+1=1;3(x + 1)2;x(x+2)=0;x1=0;x2= –2.№ 824.f(x)=(x–1)(x–2)(x–3)=(x2–3x+2)(x–3)=x3–3x2–3x2+9x+2x–6=x3–6x2+11x–6f ′(x)=3x2–12x+11, f ′(x)=11 ⇒ 3x2–12x+11=11; x(3x–12)=0; x1=0, x2=4.№ 825.1) f ′(x)=4x3–8x,f ′(x)>0, 4x3–8x>0+–4x(x2–2)>0+–02x2х∈(– 2 ; 0)∪( 2 ; +∞)2) f ′(x)=12x –12x –24xf ′(x)>0, 12x(x2–x–2)>0Решим уравнение: x(x2–x–2)=0, х=0, x2–x–2=0, D=1+8=9,1+ 31− 3х1==2, х2== –1,2232+–+–1–0х∈(–1; 0)∪(2;+∞).(x2) ()( )′′′3) f ′(x)= (x + 2 )2 x = (x + 2)2 x + (x + 2 )2 ⋅ x =1x+2(4x + x + 2) = (x + 2)(5x + 2)= 2(x + 2) x + (x + 2)2 ⋅=2 x2 x2 xx >0 ⇒ (x+2)(5x+2)>0f ′(x)>0x>0++––2−учитывая, х>025x∈(0; +∞);′′4) f ′(x)= (x − 3) x = (x − 3)′ x + (x − 3) ⋅ x =12 x + x − 3 3x − 3= x + (x − 3) ⋅==;2 x2 x2 x()( )2 x >0 ⇒ f ′(x)>0, если 3х–3>0 x>1.Учитывая, что x>0, получим х∈(1; +∞).№ 826.() ()()′′′1) f′(x)= (5 − 3x )4 (3x − 1)3 = (5 − 3x )4 (3x − 1)3 + (5 − 3x )4 (3x − 1)3 == 4 ⋅ (− 3)(5 − 3x ) (3x − 1) + 3 ⋅ 3(5 − 3x ) (3x − 1) =3342= 3(5 − 3x )3 (3x − 1)2 (− 12x + 4 + 15 − 9 x ) = 3(5 − 3x )3 (3x − 1)2 (19 − 21x )17f ′(x)<0 при 3(5 − 3x )3 (3x − 1)2 (19 − 21x ) <0.Т.к.

3>0, (3x–1)2>0, то (5 − 3x )3 (19 − 21x ) <0.++–192153⎛ 19 5 ⎞; ⎟.⎝ 21 3 ⎠Ответ: x ∈ ⎜() () ()′′′2) f′(x)= (2x − 3)2 (3 − 2 x )3 = − (2 x − 3)2 (2 x − 3)3 = − (2 x − 3)5 == −5 ⋅ 2(2 x − 3) = −10(2 x − 3) ,44f ′(0)<0 при –10(2х–3)4<0 ⇒ (2х–3)4>0 ⇒ x ≠Ответ: x ≠3.2()(3.2)′′′22⎛ 3x 2 − 1 ⎞⎟ = 3x − 1 (1 − 2x ) − 3x − 1 (1 − 2 x ) =3) f ′(x)= ⎜2⎜ 1 − 2x ⎟(1 − 2x )⎝⎠=() = 6x − 12x6 x (1 − 2 x ) + 2 3x 2 − 1(1 − 2x )22+ 6x 2 − 2(1 − 2x )f ′(x)<0 при –2(3x2–3x+1)<0 ,уравнение.2=− 6x 2 + 6x − 2(1 − 2x )23x2–3x+1>0. Решим соответствующееD=9–12<0 – нет решений, следовательно, f ′(x)<0 при всех х, кромеОтвет: х ≠1.2( )′′3′323⎛ 3x 3 ⎞⎟ = 3x (1 − 3x ) − 3x (1 − 3x ) = 9x (1 − 3x ) + 9 x =4) f ′(x)= ⎜22⎜ 1 − 3x ⎟(1 − 3x )(1 − 3x )⎝⎠=9 x 2 − 27 x 3 + 9x 3(1 − 3x )2=9x 2 − 18x 3(1 − 3x )2f ′(x)<0 если 9x2(1–2x)<0;Ответ: x >№ 827.,(1–2x)<0 ⇒ x>1.2v(t)=(ϕ(t))′=(0,1t2–0,5t+0,2)′=0,2t–0,5,v(20)=0,2⋅20–0,5=4–0,5=3,5.18121.2№ 828.v(t)=(s(t))′=(1–t+t2)′= –1+2t, v(10)= –1+2⋅10=19(м/с),mv2 5 ⋅ (19 )2==902,5 Дж.22№ 829.ρ(l)=m′(l)=(2l2+3l)′=4l+3,1) ρ(3)=4⋅3+3=15 (Г/см);2) ρ(25)=4⋅25+3=103 (Г/см).№ 830.При x<2 и x>3 подкоренное выражение положительно.1 ′−1⎛⎞ 12x − 5f ′(x)= ⎜⎜ x 2 − 5x + 6 2 ⎟⎟ = x 2 − 5x + 6 2 (2x − 5) =.2⎝⎠2 x 2 − 5x + 6()()§ 47 Производные некоторых элементарных функций№ 831.2) (еx+x2)′=ex+2x;№ 832.(′′ ⎛ 1 ⎞′1⎞1⎛3) ⎜ e2 x + ⎟ = e 2 x + ⎜ ⎟ = 2e 2 x − 2 ;x⎠x⎝⎝x⎠′′1−3x−3x ′−3x4) e+ x = e+ x = −3e+.2 x( )1) (еx+1)′=ex;() () ( ) ( )) ( )′′′1) e 2 x +1 + 2 x 3 = e 2 x +1 + 2 x 3 = 2e 2 x +1 + 6 x 2 ;′′′ 1 1 x −11⎛ 1 x −1⎞ ⎛ 1 x −1 ⎞2) ⎜ e 2−;− x −1⎟ = ⎜e 2 ⎟ − x −1 = e 22⎠⎝⎠ ⎝2 x −1′′′ ⎛ 1 ⎞⎛1 ⎞1⎟ = e 0, 3 x + 2 + ⎜⎟ = 0,3e0,3x + 2 −;3) ⎜⎜ e0,3x + 2 +⎟⎜⎟x⎠2x x⎝⎝ x⎠′′′4) e1− x + x −3 = e1− x + x −3 = −e1− x − 3x − 4 ;2 ′23 ′36) ⎛⎜ e2 x ⎞⎟ = 6 x 2 ⋅ e2 x ;5) ⎛⎜ e x ⎞⎟ = 2 x ⋅ e x ;⎝⎠⎝⎠((())) ( ) ( )№ 833.1) (2x+ex)′=2xln2+ex2) (3x–x–2)′=3xln3+2x–3 (опечатка в ответе задачника)3) (e2x–x)′=2e2x–1; 4) (e3x+2x2)′=3e3x+4x′22 ′225) ⎛⎜ 3x + 2 ⎞⎟ = ⎛⎜ 9 ⋅ 3x ⎞⎟ = 18x ⋅ 3x ⋅ ln 3 = 2 x ⋅ 3x + 2 ⋅ ln 3⎝⎠ ⎝⎠(Опечатка в ответе задачника).19№ 834.(())()′′1) 0,5x + e3x = 0,5x ln 0,5 + 3e3x ; 2) 3x − e2 x = 3x ln 3 − 2e 2 x ;′′1 ⎞41⎛; 4) ⎜ e3− x + 4 ⎟ = −e3− x − 5 .3) e 2 − x + 3 x = −e 2 − x +3 2xx⎝⎠3 x№ 835.1) (2 lnx +3x)′=23+3 x ln3; 2) (3 lnx –2x)′= +2x ln2;xx′′1 ⎞111⎛;3) ⎜ log 2 x +− 2 ; 4) 3x −3 − log3 x = −9 x − 4⎟ =x ln 32x ⎠x ln 2 2 x⎝(((6) ((3x))′ = 1 ⋅x(x −−22xx ) = x2x−−22x ;′′− 2 )log x ) = (3x − 2 ) log x + (3x25) ln x 2 − 2 x22223= 6x log3 x +=)32)− 2 (log3 x )′ =3x − 2 6 x ln x 3x − 2 3x (2 ln x + 1) − 2⋅+==x ln 3ln 3x ln 3x ln 32223 x(2 ln x + 1)2−.ln 3x ln 3№ 836.1) (sin x +х2)′=cos x +2х;3) (cos x +ех)′= –sin x +ех;2) (cos x –1)′= –sin x +0= –sin x; 4) (sin x –2х)′=cos x –2х ln 2.№ 837.1) (sin (2х–1))′=2cos (2х–1) 3) (sin (3–х))′= –cos (3–х);2) (cos (x +2))′= –sin (х+2) 4) (cos (х3))′=3х2⋅(–sin (х3))= –3х2sin x3.№ 838.1 ⎛x ⎞⎛x ⎞1) (cos ⎜ − 1⎟ + e3x )′ = − sin ⎜ − 1⎟ + 3e3x2 ⎝2 ⎠⎝2 ⎠1⎛x⎞⎛x⎞2) (sin ⎜ + 3 ⎟ + 2 x )′= cos⎜ + 3 ⎟ +2 х ln 23 ⎝3⎠⎝3⎠3) (3cos 4x–11)′= –12sin4x+ 22x2x№ 839.( )′′⎛ cos x ⎞ (cos x )′ e x − cos x ⋅ e x1) ⎜⎜ x ⎟⎟ ==e2 x⎝ e ⎠=20− sin x ⋅ e x − cos x ⋅ e xe2x=− sin x − cos xex;( )′′x ′x⎛ 3x ⎞⎟ = 3 sin x − 3 (sin x ) =2) ⎜2⎜ sin x ⎟sin x⎝⎠=3x ⋅ ln 3 ⋅ sin x − 3x cos x2sin x=3x (ln 3 ⋅ sin x − cos x )sin 2 x3) (ln x⋅cos 3x)′=(ln x)′cos 3x+lnx⋅(cos3x)′=;cos 3x+lnx⋅ 3(–sin 3x)=xcos 3x–3 ln x⋅sin 3x;x4) (log2x⋅sin 2x)′=(log3x)′sin 2x+log3x(sin 2x)′=1sin 2 x=+ sin 2 x + log 3 ⋅ 2 cos 2 x =+ 2 log 3 x cos 2 x .x ln 3x ln 3=№ 840.(2) f ′(x)= (e)′22, f ′(2)= 2e 2⋅2 − 4 + =2+1=3;2x′33x−2− ln (3 x − 1) = 3e3 x − 2 −,3x − 133⋅ 2 − 22=3–3=0;f ′( )= 3e 3 − 233 3 −11) f ′(x)= e 2 x − 4 + 2 ln x = 2e 2 x − 4 +)3) f ′(x)=(2 x–log 2 x)′=2 x ln2–1,x ln 22 ln 2 2 − 111;=2 ln2–=1 ⋅ ln 2ln 2ln 2′14) f ′(x)= log0,5 x − 3x =− 3x ln 3 ,x ln 0,511− 31 ln 3 =− 3 ln 3 .f ′(1)=ln 0,5ln 0,5f ′(1)= 21 ln2–()№ 841.1) f ′(x)=(x–cos x)′=1+sin x, f ′(x)=0 ⇒ 1+sin x=0 ⇒ sin x= –1,πx = − +2πn, n∈Z;2112) f ′(x)=( x–sin x)′= –cos x,2211f ′(x)=0 ⇒–cos x=0, ⇒ cos x = , откуда22πх= ± +2πn, n∈Z;32122 − x − 3 −x −1,−1 ==x+3x+3x+3f ′(x)=0 ⇒ –(x+1)=0, ⇒ х= –1;1–2,4) f ′(x)<(ln(x+1)–2x)′=x +1111f ′(x)=0 при–2=0, т.е.

х+1=⇒ x= – ;x +1223) f ′(x)=(2 ln(x+3)–x)′=5) f ′(x)=(x2+2x–12lnx)′=2x+2–1212, f ′(x)=0 ⇒ 2x+2–=0 x≠0,xx2x2+2x–12=0; x2+x–6=0, D=1+24=25, х1=−1 + 5−1− 5= –3.=2, х2=22Т.к. x>0 ⇒ x=2.6) f ′(x)=(x2–6x–8⋅ lnx)′=2x–6–88, f ′(x)=0 ⇒ 2x–6– =0, x≠0,xx2x2–6x–8=0; х2–3х–4=0; D=9+16=25, х1=3+53−5=4, х2== –1,22x>0 ⇒ x=4.№ 842.1) f ′(x)=(e x–x)′=e x –1, f ′(x)>0 при e x –1>0, т.е. e x >1или e x >е 0, откуда x>0;2) f ′(x)=(x ln 2–2 x)′=ln 2–2 x ln 2,++f ′(x)>0 при ln2(1–2 x)>0,xx–т.к. ln 2>0, то 1–2 >0 или 2 <1,–202x <20, откуда x<0;x 2x 2x3) f ′(x)=(e ⋅x )′= e ⋅x +2xe ,f ′(x)>0 при e x(x2+2x)>0, e x>0, x(x+2)>0Ответ: x∈(–∞; –2)∪(0;+∞).()′4) f ′(x)= e x x = e x x +ex2 x–+1x+0x−12f ′(x)>0 при e x (1+) > 0 ),2x11т.к. e x x >0, то 1+>0 ⇒ x > −f ′(x)>0 и x >0.

Ответ: х > 0.2x2x№ 843.′⎛ 2x − 12 x + 3 ⎞⎟1⋅ 22⋅512=+=+1) ⎜+ ln⎜⎟355(2x+3)2x+33⋅22−16−3xx⎝⎠′⎛ 1− x(− 1) − 2 ⋅ 3 ⋅ (− 5) = − 6 + 10 .2 − 5x ⎞⎟1=⋅2) ⎜− 2 ln⎜ 6⎟3 ⎠6 2 1 − x (2 − 5x ) ⋅ 312 1 − x 2 − 5x⎝21′1− x ⎞1− x ⎞⎛ 1− x⎛ 1 ⎞ 1− x⎛ 1 ⎞⎛3) ⎜ 2e 3 + 3 cos⎟ = 2 ⋅ ⎜ − ⎟e 3 + 3 ⋅ ⎜ − ⎟⎜ − sin⎟=2 ⎠2 ⎠⎝⎝ 3⎠⎝ 2 ⎠⎝2 1− x 31− x.= − e 3 + sin322′2− x11+ x1+ x ⎞11+ x⎛ 2− x⎛ 1 ⎞ 2− x= − e 3 − cos4) ⎜ 3e 3 − 2 sin.⎟ = 3 ⋅ ⎜ − ⎟e 3 − 2 ⋅ cos4 ⎠4424⎝⎝ 3⎠№ 844.′⎛1x − 2 ⎞⎟ ⎛ 3x−2⎞3⎜3= ⎜ 3 ⋅ (2 − x )− 3 − 3 cos1)− 3 cos⎟=⎜ 2− x⎟ ⎝33 ⎠⎝⎠4x−2⎞1⎛x−21⎛ 1⎞= 3 3 ⋅ ⎜ − ⎟ ⋅ (− 1) ⋅ (2 − x )− 3 − 3 ⋅ ⎜ − sin.+ sin⎟=33⎝3 ⎠ (2 − x )3 (2 − x ) ⋅ 9⎝ 3⎠′x−4 ⎞⎛⎛ ⎛ 3⎞1 x−4 ⎞1−75 ⎟ = ⎜2⋅ −−2) ⎜ 2 ⋅ 4e5⎜⎟(x + 2 ) 4 − 5 ⋅ ⋅ e 5 ⎟⎟ =⎜3⎜⎟ ⎝ ⎝ 4⎠5(x + 2 )⎠⎝⎠x−473= − (x + 2)− 4 − e 5 .2№ 845.() ( )′′1) 0,5 x ⋅ cos 2 x = 0,5 x cos 2 x + 0,5 x (cos 2 x )′ == 0,5 ln 0,5 cos 2 x + 0,5 x ⋅ 2 ⋅ (− sin 2 x ) = 0,5 x (ln 0,5 cos 2 x − 2 sin 2 x ) .′′′⎞⎛2) 5 x ⋅ e− x = 5⎜ x e − x + x e− x ⎟ =⎝⎠x()( )( )⎞⎛ 1⎞ 5e− x (1 − 2 x )⎛ 1 −x.= 5⎜⎜e − x ⋅ e − x ⎟⎟ = 5e − x ⎜⎜− x ⎟⎟ =2 x⎝2 x⎠⎝2 x⎠3) (e3–2 x⋅cos(3–2x))′=(e3–2 x)′cos(3–2x)+ e3–2 x(cos(3–2x))′== –2 e3–2 xcos (3–2x)–2 e3–2 x⋅(–sin(3–2x))==–2e3–2 x(cos(3–2x)–sin(3–2x)) = 2e3–2 x(sin(3–2x)–cos(3–2x)).№ 846.()′1) ln x − 1 =1⋅1=1.2( x − 1)x −1 2 x −1′12) ⎛⎜ e 3+ x ⎞⎟ =e 3+ x (опечатка в ответе задачника).⎝⎠2 3+ x113) (ln (cos x ))′ =⋅ (–sin x)= –tg x.

4) (ln (sin x ))′ =⋅ cos x =ctg x.cos xsin x22№ 847.( )′3) (cos x + 2 ) = − sin()′′1) 2cos x +1 = 2cos x +1 ⋅ln 2⋅(–sin x). 2) 0,51+ sin x = 0,51+ sin x ⋅ln0,5⋅cos x.33x+2⋅331(x + 2)− 2 = − sin x + 2 .333 x + 22()1 cox(ln x)4) (sin (ln x))′=cos(ln x)⋅ =.xx№ 848.′(2 x + 2) =x +1⎛⎞1) ⎜ x 2 + 2 x − 1 ⎟ =22⎝⎠2 x + 2x − 1x + 2x − 1′′ 1 ⋅ (− sin x)sin xcos x34. 3) cos x = 4=− 4.2) sin x =34 cos x4 cos x3 sin 2 x′114) log 2 x ==2 log 2 x ⋅ x ln 2 2 x ln 2 ⋅ log 2 x()(())№ 849.′′⎛ 1 + cos x ⎞ (1 + cos x ) sin x − (1 + cos x )(sin x)′1) ⎜=⎟ =sin 2 x⎝ sin x ⎠=− sin x ⋅ sin x − (1 + cos x ) ⋅ cos x2=32 3x− sin 2 x − cos x − cos 2 x( 3x )(′ 3 + 1)− 3x (3 + 1)′ =(3 + 1)(3 + 1)− 3x (3 ln 3)=(3 + 1)′⎛ 3x ⎞⎟ =2) ⎜ x⎜ 3 +1⎟⎝⎠=sin x=x=−1 + cos xsin 2 xxxxsin 2 x2xx23 ⋅ 3x + 3 − 2 ⋅ 3x ⋅ 3x ln 3=3 x +1 (1 − 2 x ln 3) + 3.22 3x 3x + 12 3x 3x + 1′′⎛ e 0, 5 x ⎞e0,5 x (cos 2 x − 5) − e0,5 x (cos 2 x − 5)′⎜⎟==3)⎜ cos 2 x − 5 ⎟(cos 2 x − 5)2⎝⎠=()( )20,5e0,5 x (cos 2 x − 5) + 2e0,5 x sin 2 x(cos 2 x − 5)2( )(=)0,5e0,5 x (cos 2 x − 5 + 4 sin 2 x )( )(cos 2 x − 5)2.′′⎛ 52 x ⎞52 x (sin 3x + 7 ) − 52 x (sin 3x + 7 )′⎜⎟4)==⎜ sin 3x + 7 ⎟(sin 3x + 7 )2⎝⎠=2 ⋅ 52 x ⋅ ln 5(sin 3x + 7) − 52 x ⋅ 3 ⋅ cos 3x(sin 3x + 7)2=52 x (2 ⋅ ln 5 ⋅ sin 3x + 14ln 5 − 3cos3x)(sin 3x + 7)223№ 850.(=(ex) ()′x−x ′⎞x − e x − e− x (x )′⎟ = e −e=⎟x2⎠⎛ e x − e− x1) ⎜⎜x⎝) ()+ e− x ⋅ x − e x − e− x ⋅1x2=e x ⋅ x + e− x ⋅ x − e x + e− x(x2)e x ( x − 1) + e x ( x + 1)=()x ⋅ 2 x ln 2 −1 − 2xln 222x2′′⎛ 2 x − log 2 x ⎞2 x − log 2 x ⋅ ln 2 ⋅ x − 2 x − log 2 x (ln 2 ⋅ x )′⎟⎜2)==⎜ ln 2 ⋅ x ⎟x 2 ⋅ ln 2 2⎠⎝=(2xln 2 −1x ln 2)⋅ ln 2 ⋅ x − (22ln 2 ⋅ xx)− log 2 x ⋅ ln 22=+ log 2 xx ⋅ ln 2№ 851..′(sin x − cos x )′ ⋅ x − (sin x − cos x ) ⋅ (x )′ =⎛ sin x − cos x ⎞1) ⎜⎟ =xx2⎝⎠(cos x + sin x ) ⋅ x − (sin x − cos x ) ⋅ 1 ==x2cos x ⋅ x + sin x ⋅ x − sin x + cos x cos x( x + 1) + sin x( x − 1)==x2x2′′2⎛ 1 − sin 2 x ⎞ ⎛⎜ (sin x − cos x ) ⎞⎟2) ⎜= (sin x − cos x )′ = cos x + sin x .⎟ =⎝ sin x − cos x ⎠ ⎜⎝ sin x − cos x ⎟⎠№ 852.()′1) f ′( x) = 5(sin x − cos x ) + 2 cos 5 x =()= 5(cos x + sin x ) + 2 ⋅ 5 ⋅ (− sin 5 x ) = 5 cos x + sin x − 2 sin 5 x =⎛⎞⎛π⎞= 5⎜⎜ cos x + cos⎜ − x ⎟ − 2 sin 5 x ⎟⎟ =2⎝⎠⎝⎠⎛⎞⎛ ⎛⎞π⎞ππ⎞⎛= 5⎜⎜ 2 sin ⎜ x + ⎟ sin − 2 sin 5 x ⎟⎟ = 5 2 ⎜⎜ sin ⎜ x + ⎟ − sin 5 x ⎟⎟ =444⎝⎠⎠⎝⎠⎝ ⎝⎠x+π4− 5xx+π4− 5x⎞⎛π⎞ ⎛π= 10 2 sin ⎜ − 2 x ⎟ cos⎜ + 3x ⎟⎠⎝8⎠ ⎝8⎛π⎞ ⎛π⎞f′(x)=0 при sin⎜ − 2 x ⎟ cos⎜ + 3 x ⎟ =0,⎝8⎠ ⎝8⎠= 5 2 ⋅ 2 sin2cos2π π⎡π⎡⎢ 8 − 2 x = πn⎢ x = 16 + 2 n, n ∈ ZОткуда: ⎢⇒⎢⎢ π + 3 x = π + πk⎢ x = π + π k, k ∈ Z⎢⎣8 32⎣⎢ 824;2) f ′(( x) = (1 − 5 cos 2 x + 2(sin x − cos x ) − 2 x )′ =⎛π⎞= 10 sin 2 x + 2(cos x + sin x ) − 2 = 10 cos⎜ − 2 x ⎟ + 2(cos x + sin x ) − 2 =⎝2⎠⎛⎛π⎞⎛π⎞⎞= 10⎜⎜ cos2 ⎜ − x ⎟ − sin 2 ⎜ − x ⎟ ⎟⎟ + 2(cos x + sin x ) − 2 =44⎝⎠⎝⎠⎠⎝⎛⎛π⎞ ⎞⎛π⎞= 10⎜⎜ 2 cos2 ⎜ − x ⎟ − 1⎟⎟ + 2 2 cos⎜ − x ⎟ − 2 =⎝4⎠ ⎠⎝4⎠⎝⎛π⎞⎛π⎞= 20 cos2 ⎜ − x ⎟ + 2 2 cos⎜ − x ⎟ − 12⎝4⎠⎝4⎠⎛π⎞⎛π⎞f ′(x)=0, если 20 cos 2 ⎜ − x ⎟ + 2 2 cos⎜ − x ⎟ − 12 =0⎝4⎠⎝4⎠πcos( –x)=t, 20t2+2 2 t–12=0 или 10t2+ 2 t–6=0 D=2+240=242=2⋅1214t1=− 2 + 11 22=202t2=− 2 − 11 23 2=−2059⋅23 2∨ 1; 18 < 25 ⇒<12553 2∨ 1;5π3 23 2>–1.)+2πn,− х1=±arccos(–554⎛ 3 2⎞π⎟ + 2πn, n ∈ Zn∈Z ⇒ x1 = ± arccos⎜ −⎜45 ⎟⎠⎝Cледовательно, –πππ− x2=± +2πk, k∈Z ⇒ x2 = 2πk , k ∈ Z , x3 = + 2πm, m ∈ Z442№ 853.() ( )′′1) f ′( x) = e 2 x ln (2 x − 1) = e 2 x ⋅ ln (2 x − 1) + e2 x (ln (2 x − 1))′ == 2e 2 x ln (2 x − 1) +2x2e1 ⎞⎛= 2e 2 x ⎜ ln (2 x − 1) +⎟,2x − 12x − 1 ⎠⎝f(x)=0 ⇒ e2xln(2x–1)=0, e 2x>0, так что ln(2x–1)=0, ln(2x–1)=ln 1;1 ⎞⎛2x–1=1; x=1, f ′(1)= 2e 2⋅1⎜ ln (2 ⋅ 1 − 1) +⎟ =2e2⋅(0+1)=2e2.⋅−1⎠21⎝′(sin x − cos x )′ sin x − (sin x − cos x )(sin x )′ =⎛ sin x − cos x ⎞2) f ′( x) = ⎜⎟ = =sin xsin 2 x⎝⎠25==(cos x + sin x )sin x − (sin x − cos x )cos x =sin 2 x2cos x ⋅ sin x + sin x − sin x cos x + cos 2 x2sin x⎛ sin x − cos x ⎞f(x)=0 ⇒ ⎜⎟=0sin x⎝⎠=1sin 2 xОбласть определения функции sin x≠0 x≠πn, n∈Z11π⎛π⎞= 1 =2.1–ctg x=0 ctg x=1 x= +πn, n∈Z ; f ′⎜ ⎟ =4⎝ 4 ⎠ sin 2 π()42Т.к.