Ответ(ы) на вопрос:
Гость1) f(x)=2x²√x-4x+11+3/(∛x)+1/x=2x^(5/2)-4x+11+3·x^(-1/3)+x^(-1) x0=1
f'(x)=2·(5/2)x^(3/2)-4+3·(-1/3)x^(-4/3)-x^(-2)=5·x^(3/2)-4 -x^(-4/3)-x^(-2)
f'(1)=5·(1)^(3/2)-4 -(1)^(-4/3)-(1)^(-2)=5-4-1-1=-1
2)f(z)=(z-1)√(z²-1)
f'(z)=(z-1)'√(z²-1) +(z-1)(√(z²-1) )'=√(z²-1) +{(z-1)/(2√(z²-1) )}·2·z=
=√(z²-1) +(z²-z)/(√(z²-1) )
z0=2
f'(2)=√(2²-1) +(2²-2)/(√(2²-1) )=√3+(2/√3)=5/√3=5√3/3
3) f(x)=√(x²-1)/x f'(x)={[√(x²-1)]'·x-(x)'√(x²-1)}/x²={2x/(2√(x²-1))·x-√(x²-1)}/x²=
={x²/(√(x²-1))-√(x²-1)}/x²=(x²-x²+1)/(x²√(x²-1))=1/(x²√(x²-1))
f'(√5)=1/[(√5)²√((√5)²-1)]=1/(5·2)=1/10=0,1
4) f(x)=[e^(2x)-e^(-2x)]/2
f'(x)=[(e^(2x))'-(e^(-2x))']/2 =[2·e^(2x)-(-2)e^(-2x)]/2 = [e^(2x)+e^(-2x)]
x0=0 f'(0)=[e^(2·0)+e^(-2·0)]=1+1=2
5)f(x)=ln[(x+1)/x] f'(x)=(x/(x+1)) ·(-1/x²)= -1/[x(x+1)]
x0=3
f'(3)=-1/[3(3+1)]= -1/12
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