Упр.876 ГДЗ Алимов 10-11 класс (Алгебра)

1) f(x) = cosxsinx, x0=пи/6;
2) f(x) = exlnx, x0=1;
3) f(x) = 2cosx/sinx, x0=пи/4;
4) f(x) = x/(1+ex), x0=0.

1. $$f(x)=\cos x\sin x,\quad x_0=\frac{\pi}{6}$$

   $$f(x)=\frac12\sin 2x$$

   $$f'(x)=\frac12\cdot 2\cos 2x=\cos 2x$$

   $$f’\!\left(\frac{\pi}{6}\right)=\cos\frac{\pi}{3}=\frac12$$

2. $$f(x)=e^x\ln x,\quad x_0=1$$

   $$f'(x)=(e^x)’\ln x+e^x(\ln x)’=e^x\ln x+\frac{e^x}{x}=e^x\left(\ln x+\frac1x\right)$$

   $$f'(1)=e\left(\ln 1+\frac11\right)=e(0+1)=e$$

3. $$f(x)=\frac{2\cos x}{\sin x},\quad x_0=\frac{\pi}{4}$$

   $$f(x)=2\ctg x$$

   $$f'(x)=2(\ctg x)’=-\frac{2}{\sin^2 x}$$

   $$f’\!\left(\frac{\pi}{4}\right)=-\frac{2}{\sin^2\frac{\pi}{4}}=-\frac{2}{\left(\frac{\sqrt2}{2}\right)^2}=-4$$

4. $$f(x)=\frac{x}{1+e^x},\quad x_0=0$$

   $$f'(x)=\frac{(1+e^x)\cdot 1-x\cdot e^x}{(1+e^x)^2}=\frac{1+e^x-xe^x}{(1+e^x)^2}$$

   $$f'(0)=\frac{1+e^0-0\cdot e^0}{(1+e^0)^2}=\frac{1+1}{(1+1)^2}=\frac{2}{4}=\frac12$$

Ответ

1) $$\frac12$$; 2) $$e$$; 3) $$-4$$; 4) $$\frac12$$.