найти производную y=arctg (3x+x^2)

Question

Алгебра: найти производную y=arctg (3x+x^2)

pokin45 · Accepted Answer

1)y=x^2 /(x+5); x∈ [-4;1]
y=f(x); f(-4)=16/(-4+5)=16/1=16; наибольшее
f(1)=1/(1+5)=1/6;
y'=(x^2 /(x+5)'=(2x(x+5)-x^2)/ (x+5)^2=(x^2+10x)/ (x+5)^2;
y'=0; x^2+10x=0; x≠-5
x(x+10)=0; x=0 ili x=-10; -10∉[-4;1]
f(0)=0/(0+5)^2=0 наименьшее
2)y=sin2x -x; [-π/2;π/2]
f(-π/2)=sin(-π) +π/2=-sinπ +π/2=π/2=1,57; наибольшее
f(π/2)=sinπ -π/2=-π/2=-1,57 наименьшее
y'=(sin2x -x)'=2cos2x -1;
y'=0; 2cos2x -1=0; cos2x=1/2; 2x=+-π/3+2πn; x=+-π/6; x∈[/π/2; π/2]!
f(-π/6)=-sinπ/3) +π/6=√3/2 +π/6≈0,85+0,53=1,38;
f(π/6)=sinπ/3-π/6=√3/2 -π/6≠0,85-0,53=0,32

MOGZ ответил