Решите уравнение:

1. f'(x)·g'(x)=0

f(x)=x³-3x²

g(x)=2/3√x

2. f(x)=(1+2x)(2x-1)

f'(-2)

3.φ(x)=3+x/√x

φ'(8)

4. g(x)=4sinx

g'(-π/3)

1

Ответы и объяснения

2012-04-05T15:18:26+04:00

1) f'(x)*g'(x) = 0

f'(x) = 3x^2-6x 

g'(x) = -\frac{1}{3\sqrt{x^3}} 

(3x^2-6x)(-\frac{1}{3}\frac{1}{\sqrt{x^3}})=0 

\frac{6x-3x^2}{3x^{\frac{3}{2}}}=0 

\frac{3x(2-x)}{3x\sqrt{x}} =0  

x\neq0 

x = 2 

2) f(x) = (1+2x)(2x-1) = 4x^2-1

f'(x) = 8x 

f'(-2) = -16 

3) f(x) = 3 + \frac{x}{\sqrt{x}} = 3+\sqrt{x}

f'(x) = -\frac{1}{2\sqrt{x}} 

f'(8) = -\frac{1}{2\sqrt{8}}=-\frac{1}{4\sqrt{2}} 

4) g(x) = 4sinx

g'(x) = 4cosx

g'(-\frac{\pi}{3}) = 4cos(-\frac{\pi}{3}) = 4*\frac{1}{2} = 2