Ответы и объяснения

Лучший Ответ!
2013-05-11T04:18:51+04:00

Немного теории.

(x^n)'=n*x^{n-1}\\(cosx)'=-sinx\\(f*g)'=f'*g+f*g'\\(\frac{f}{g})'=\frac{f'*g-f*g'}{g^2}\\cos0=1\\S'(t)=v(t)\\(c)'=0\\(\sqrt{f})'=\frac{f'}{2*\sqrt{f}}\\k=f'(x)

Применяем.

№6f'(x)=(x^2-x)'*cosx+(x^2-x)*(cosx)'=\\=(2x-1)*cosx+(x^2-x)*-sinx\\f'(x_{0})=f'(0)=(2*0-1)*cos0+(0^2-0)*-sin0=\\=-1*1+0=-1

№7f'(x)=\frac{(x^2+2)'*(x)-(x^2+2)*(x)'}{x^2}=\frac{2x*x-x^2-2}{x^2}=\frac{x^2-2}{x^2}

       g'(x)=6+\frac{(2)'*(x)-(2)*(x)'}{x^2}=6+\frac{0-2}{x^2}=6-\frac{2}{x^2}

       \frac{x^2-2}{x^2}= 6-\frac{2}{x^2}\ \ \ \ |*x^2\\x^2-2=6x^2-2\\5x^2-2+2=0\\5x^2=0\\x=0

№8S'(t)=4t-20=v(t)

v(t)=8\\4t-20=8\\4t=28\\t=7

№9 Пусть угловой коффициент касательной равен k. k=f'(x) 

f'(x)=\frac{(x^2-4x-1)'}{2\sqrt{x^2-4x-1}}=\frac{2x-4}{2\sqrt{x^2-4x-1}}\\f'(x_{0})=f'(5)=\frac{10-4}{2\sqrt{25-20-1}}=\frac{6}{2*2}=\frac{6}{4}=\frac{3}{2}

2013-05-11T04:54:21+04:00

1.) (uv)'=u'v+uv'\\\\ (u-v)'=u'-v'\\\\ x^{n}=n\cdot x^{n-1}\\\\ x'= 1\\\\ cosx'=-sinx\\\\ f(x)=(x^{2}-x)\cdot cosx\\ f'(x)=((x^{2}-x)\cdot cosx)'=(x^{2}-x)'\cdot cosx + (x^{2}-x)\cdot cosx'=\\ =(2x-1)\cdot cosx+(x^{2}-x)\cdot(-sinx)\\\\ f'(0)=(2\cdot0-1)\cdot1+(0^{2}-0)\cdot0=-1

 

2.) f(x)=\frac{x^{2}+2}{x}\\ g(x)=6x+\frac{2}{x}\\\\ f'(x)=(\frac{x^{2}+2}{x})'=\frac{(x^{2}+2)'x-(x^{2}+2)x'}{x^{2}}=\frac{2x^{2}-x^{2}-2}{x^{2}}=\frac{x^{2}-2}{x^{2}}\\\\ g'(x)=(6x+\frac{2}{x})'=(6x)'+(\frac{2}{x})'=(6'x+6x')+(\frac{2'x-2x'}{x^{2}})=\\ =6-\frac{2}{x^{2}}\\\\ f'(x)=g'(x)\\\\ 6-\frac{2}{x^{2}}=\frac{x^{2}-2}{x^{2}} \ | \ \cdot x^{2}\\ 6x^{2}-2=x^{2}-2\\ 6x^{2}-x^{2}-2+2=0\\ 5x^{2}=0\\ x=0 

 

3.) S(t)=2t^{2}-20t+3\\\\ v(t)=S'(t)\\\\ v(t)=4t-20\\ 4t-20=8\\ 4t=8+20\\ 4t=28 \ | \ : 4\\ t=7

 

Ответ: при t = 7 секунд.

 

4.) y'=(\sqrt{u})'=\frac{u'}{2\sqrt{u}}\\\\ f(x)=\sqrt{x^{2}-4x-1} \ ; \ x_{0}=5\\\\ k = f'(5)\\\\ f'(x)=\frac{2x-4}{2\sqrt{x^{2}-4x-1}}\\\\ f'(x_{0})=f'(5)\frac{2\cdot 5-4}{2\sqrt{5^{2}-4\cdot 5-1}}=\frac{6}{2\sqrt{25-20-1}}=\frac{6}{2\sqrt{4}}=\frac{6}{4}=\frac{3}{2}