Задача по "Математике"

Задание 1.

1. =1⁴ -3*1+2/1⁵-4*1+3=[0/0]=lim 4x³-3/5x⁴-4=4*1³-3/5*1⁴-4=4-3/5-4=1/1=1

 

2. =[∞/∞]=(выносим  х² из числителя и знаменателя за скобки 1/∞=0)=lim x²(1+2/x+3/x²)/x²(-2+3/x+4/x²)=lim 1/2=1/2

 

3. =∞+3/2*∞²+3*∞+4=[∞/∞](по правилу Лопиталя)=lim 1/2*2x+3=1/4*∞+3=1/∞=0

 

4. =К(1+0)-1/0=1-1/0=[0/0](домножить числитель и знаменатель на корень(1+х)+1)=lim (К(1+х)-1)(К(1+х)+1)/х(К(1+х)+1)={(a+b)(a-b )=a²-b²}=lim (К1+х)²-1²/х(К(1+х)+1)=lim1+x-1/x(К(1+х)+1)=lim x/x(К(1+х)+1)=lim 1/К(1+х)+1=1/К(1+0)+1=1/1+1=1/2

 

5. 1³+4*1-1/3*1²+1+2=4/6=2/3

 
 

6. =∞-к∞²+∞+1=[∞-∞]={домножим и разделим на х+Кх²+х+1}=lim (x-Кх²+х+1)(х+ Кх²+х+1)/х+ Кх²+х+1={(a-b)(a+b)= a²-b²}=lim x²-(x²+x+1)/x+Кх²+х+1=lim  –x-1/ x+Кх²+х+1= -∞-1/∞+К∞²+∞=[∞/∞]={выносим Х за скобки}=lim x(-1-1/x)/x(1+К1+1/х+1/х²)=lim -1/1+К1=-1/2

 
 

7. =sin5*0-sin3*0/0=[0/0]=lim sin5x/x-lim sin3x/x=lim 5x/x-lim 3x/x=lim 5-lim3=5-3=2

 

8. =Кв3степени(1+0²)-1/К(3*0²+1)-1=1-1/1-1=[0/0]={(a-b)(a²+ab+b²)=a³-b³}=lim (Кв3степени(1+х²)-1)((Кв3степени(1+х²))²+ Кв3степени(1+х²)+1)/ (К(3х²⁺1-1)((Кв3степени(1+х²))²+ Кв3степени(1+х²)+1)=lim (Кв3степ(1+х²)³)-1³/(К(3х²⁺1-1)((Кв3степени(1+х²))²+ Кв3степени(1+х²)+1)=lim 1+x²-1/(К(3х²⁺1-1)((Кв3степени(1+х²))²+ Кв3степени(1+х²)+1)=lim x²/(К(3х²⁺1-1)((Кв3степени(1+х²))²+ Кв3степени(1+х²)+1)=[0/0]= ={(a-b)(a+b)= a²-b²}=lim x²((K3x²+1)+1)/((K3x²+1)-1)((K3x²+1)-1)((Кв3степ1+х²)²+(Кв3степ1+х²)+1)=lim x²((K3x²+1)+1)/(3x²+1-1)((K3x²+1)²+(Кв3степ1+х²)+1)=lim x²((K3x²+1)+1)/3x²((K3x²+1)²+(Кв3степ1+х²)+1)= (K3*0²+1)+1/3((Kв3степ1-0²)²+(Кв3степ1+0²)+1=1+1/3((Kв3степ1)²+Кв3степ1+1=2/3(1+1+1)=2/9

 

9. =(К1+0+0²)-1/3*0=1-1/0=[0/0]= {(a-b)(a²+ab+b²)=a³-b³}=lim ((K1+x+x²)-1)((K1+x+x²)+1)/3x((K1+x+x²)+1)=lim (K1+x+x²)²-1/3x((K1+x+x²)+1)=lim 1+x+x²-1/3x((K1+x+x²)+1)=lim x(x-1)/3x((K1+x+x²)+1)= lim x+1/3x((K1+x+x²)+1)=0+1/3((K1+0+0²)+1)=1/3((K1)+1)=1/3*2=1/6

 
 

10. =[0/0]= {a³-b³=(a-b)(a²+ab+b²)}=lim (x-1)(x²+x+1)/x-1=lim (x²+x+1)=1²+1+1=3

 
 

11. =(∞+5/∞+1)^∞=[ ∞/∞]^∞=lim (x+1+4/x+1)^x=lim (x+1/x+1+4/x+1)^x=lim(1+4/x+1)^x=(1+4/∞+1)^∞=[1^∞]={limf(x)^φ(x)=[1^∞]=lim [f(x)-1]*φ(x)}=lim [1+(4/x+1)-1]*x=lim 4x/x+1=по правилу Лопиталя=lim 4/1=i⁴

 

12. =[0/0]{lim e^x-1~x}=lim 3x²/5x²=3/5

 

13. =lim -9x²/x²(1+3/x²)= -9/1= -9

 
 

 

Задание 2

1. y^’=(log₂x+3ln(x²+5))^’={(a+b)^’=a^’+b^’}=(log₂x)^’+(31n(x²+5))^’={(log_ax^’)=1/x+1n a;(ln u)^’=(1/u)*u^’}=1/x*ln2+3*1/x²+5*(x²+5)^’={(xⁿ)^’=nx^n-1;с^’=0(с постоянное число)}=1/x*ln2+3*1/x²+5*(2x+0)=1/xln2+6x/x²+5

 

2. y^’=(Кв7степ x+2)^’={КвМстеп х=х*1/m}=((x+2)^1/7)^’={(uⁿ)^’=n*u^n-1*u^’}=1/7(x+2)^(1/7)-1*(x+2)^’=(1/7)*(x+2)^6/7*(1+0)={x^-n/m=1/КвМстеп хⁿ}=1/7*1/Кв7степ(х+2)⁶

 
 

3. y^’=(lnК х+1)^’={(ln u)^’=1/u*u^’}=1/K x+1*(K x+1)^’-{(K u)^’=1/2K u*u^’}=1/K x+1*1/2K x+1*(x+1)^’=1/2(x+1)*(1+0)=1/2(x+1)

 
 

4. y^’=(K2x-sin2x)^’={(K u)^’=1/2K u*u^’}=1/2K2x-sin2x*(2x-sin2x)^’={(sin u)^’=cos u*u^’}=1/2(K2x-sin2x)*(2*1-cos2x(2x)^’)=1/2(K2x-sin2x)*(2-2*cos2x)=2(1-cos2x)/2K2x-sin2x=1-cos2x/K2x-sin2x

 

5. y^’=(x*arctgK7+x²)^’={(a*b)^’=a^’b+ab^’}=x^’*(arctgK7+x²)^’={(arctg)^’=1/1+u²*u^’}=1*arctg(K7*x²)+x*1/1+(K7+x²)²*(K7+x²)^’={(Ku)^’=1/2(K u)*u^’}=arctg(K7+x²)+x*1/1+(7+x²)*1/2(K7+x²)*(7+x²)=arctg(K7+x²)+x*(1/8+x²)*(1/2K7+x²)*(0+2x)=arctg(K7+x²)+2x^2x/2((8+x²)*K7+x²)=arctg (K7+x²)+x²/((8+x²)K7+x²)

 
 

6. y^’=(x*tgx/1+x²)^’={(a/b)^’=a^’b-ab^’/b²}=(x*tgx)^’(1+x²)-x*tgx*(1+x²)^’/(1+x²)²={(a*b)^’=a^’b+ab^’}=(x^’*tgx+x*(tgx)^’)(1+x²)-x*tgx*(0+2x)/(1+x²)²={(tgx)^’=1/cos²x}=(1*tgx+x*1/cos²x)(1+x²)-2x²*tgx/(1+x²)²=(tgx+x/cos²x)(1+x²)-2x²*tgx/(1+x²)²

 

7. y^’=(ln(x⁶/3-x²))^’={(ln u)^’=1/u*u^’}=1/(x⁶/3-x²)*(x⁶/3-x²)^’={(a/b)^’=a^’b-ab^’/b²}=(3-x²/x⁶)*((x⁶)*(3-x²)-x^6x(3-x)^’)/(3-x²)²=6x⁵(3-x²)-x⁶(0-2x)/x⁶(3-x²)=x⁵(6(3-x²)-x(-2x))/x⁶(3-x²)=18-6x²+2x²/x(3-x²)=18-4x²/x(3-x²)

 
 

8. y^’=(6tgx/K x)^’=(6*tgx)^’*(Kx)-6tgx*(Kx)^’/(Kx)²={(tgx)^’=1/cos²x}=6*(1/cos²x)*Kx-6tgx*(1*2Kx)/x=(6*(1/cos²x)*(Kx)-3tgx*(1/Kx))/x

 

9. y^’=(tgx+(K1+x²)/x)^’=(tgx+(k1+x²))^’*x-(tgx+K1+x²)*x^’/x²=(1/cos²x+1/2(K1+x²))^’x-(tgx+K1+x²)*1/x²=(1/cos²x+1/2(1+x²)*2x)x-tgx-K1+x²/x²=((1/cos²x)+x/K1+x²)x-tgx-(K1+x²)/x²

 
 
 

10. y^’=(arcctg²1/x)^’={(arcctgx)^’=1/1+x²}=2*arcctg1/x*(arcctg1/x)^’=2*arcctg1/x*(-1/1+(1/x)²)*(1/x)^’={(1/x)^’= -1/x²}= -2arcctg 1/x*1/(1+1/x²*(-1/x²))=2arcctg1/x*x²/(x²+1)*1/x²=(2arcctg1/x)/x²+1

 
 

11. y=1/5ctgKx+6x²

y^’=(1/5ctgKx+6x²)^’={(ctg u)^’= (-1/sin²u)*u^’}= -1/5*(1/sin²Kx+6x²)*(Kx+6x²)^’= -1/5*(1/sinКв2степ x+6x²)*(1/2Kx+6x²)*(x+6x²)^’= (-1/10)*(1/sin²Kx+6x²)*(1/Kx+6x²)*(1+12x) 

12. y^’=(tg(x+x⁷))^’=(1/cos²(x+x⁷))*(x+x⁷)^’=(1/cos²(x+x⁷))*(1+7*x⁶)=1+7x⁶/cos²(x+x⁷)

 
 
 

13. y^’=(x³+ln(x+1))^’=(x³)^’+(ln(x+1))^’=3x²+(1/x+1)*(x+1)^’=3x²+(1/x+1)*(1+0)=3x²+1/x+1

 
 

14. y^’=(Kx⁶-cosx)^’=(1/2Kx⁶-cosx)*(x⁶-cosx)^’={(cosx)^’= -sinx}=(1/2Kx⁶-cosx)*(6x⁵-(-sinx))=6x⁵+sinx/2Kx⁶-cosx

 

15. y^’=(5arcctg(x³+3))^’={(arcctg u)^’= -(1/1+u²)*u^’=(-1/1+u²)*u^’}=5*(-1/1+(x³+3)²)*(x³+3)^’= -5*(1/1+(x³+3))²*(3x²+0)= -15x²/1+(x³+3)²

 

 

16.